MOTION IN A STRAIGHT LINE MCQs for JEE PHYSICS — Easy

1 Easy

1 of 31

Application +4 | -1

A ball is thrown vertically upward with speed 15 m/s and it returns to the ground with speed 10 m/s as a constant air resistance acts. The maximum height attained by the ball is

View answer and explanation

Correct option: A

11.25

Option A (11.25) is correct because it is calculated using kinematic equations, specifically the equation relating final velocity, initial velocity, acceleration, and displacement.

Solution approach

Correct Answer + Core Reason: 11.25. The maximum height is calculated using kinematic equations.

Step-by-Step Solution:

Step 1: Identify the given information: Initial velocity $u = 15 \mathrm{~m/s}$ , final velocity at maximum height $v = 0 \mathrm{~m/s}$ , acceleration due to gravity $g = 10 \mathrm{~m/s^2}$ .

Step 2: Recall the kinematic equation: ${v^2} = {u^2} + 2as$ , where $s$ is the displacement (height).

Step 3: Substitute the given values into the formula: ${0^2} = {15^2} + 2(-10)s$

Step 4: Solve for $s$ : $0 = 225 - 20s$
$20s = 225$
$s = \frac{{225}}{{20}} = 11.25 \mathrm{~m}$

Final Answer: $s = 11.25 \mathrm{~m}$

Why other options are wrong:
- Option B (10.25): This value is incorrect. It might result from a calculation error or using an incorrect formula.
- Option C (9.25): This value is incorrect. It might result from a calculation error or using an incorrect formula.
- Option D (10.5): This value is incorrect. It might result from a calculation error or using an incorrect formula.

Key Concepts:
- Kinematics
- Equations of motion

Memory Aids:
- Remember the equations of motion: $v = u + at$ , $s = ut + \frac{1}{2}a{t^2}$ , ${v^2} = {u^2} + 2as$ .

Quick Tips:
- Choose the appropriate kinematic equation based on the given information.
- Pay attention to the signs of velocity and acceleration.

2 Easy

2 of 31

Application +4 | -1

An object is projected upwards with a velocity of 100m/s. It will strike the ground after (approximately)

View answer and explanation

Correct option: B

20 sec

Option B (20 sec) is correct because it is the time of flight calculated using the formula

T = \frac{2u}{g}

.

Solution approach

Correct Answer + Core Reason: 20 sec. The time of flight is calculated using the initial velocity and acceleration due to gravity.

Step-by-Step Solution:

Step 1: Identify the given values: $u = 100 \text{{ m/s}}$ , $g = 10 \text{{ m/s}}^2$ (approximate).

Step 2: Use the formula for the time of flight: $T = \frac{{2u}}{g}$

Step 3: Substitute the values: $T = \frac{{2 \times 100 \text{{ m/s}}}}{{10 \text{{ m/s}}^2}}$

Step 4: Calculate the time of flight: $T = 20 \text{{ s}}$

Why other options are wrong:
- Option A (10 sec): This is incorrect. It is the time to reach the maximum height, not the total time of flight.
- Option C (15 sec): This is incorrect. It might be a calculation error.
- Option D (5 sec): This is incorrect. It might be a calculation error.

Key Concepts:
- Projectile motion
- Time of flight

Memory Aids:
- $T = \frac{{2u}}{g}$

Quick Tips:
- Remember the formula for the time of flight.
- Use $g = 10 \text{{ m/s}}^2$ for approximations.

3 Easy

3 of 31

Understanding +4 | -1

Find the average velocity when a particle complete the circle of radius

1 \mathrm{~m}

in

10 \mathrm{~s}

.

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because the average velocity is zero when the particle completes a full circle, as the displacement is zero.

Solution approach

Correct Answer + Core Reason: Zero. The average velocity is zero because the displacement is zero after completing the circle.

Step-by-Step Solution:

Step 1: Define average velocity: Average velocity is defined as the total displacement divided by the total time taken: $v_{avg} = \frac{\Delta x}{\Delta t}$ , where $\Delta x$ is the displacement and $\Delta t$ is the time interval.

Step 2: Determine the displacement: When a particle completes a circle, its final position is the same as its initial position. Therefore, the displacement is zero: $\Delta x = 0$ .

Step 3: Calculate the average velocity: $v_{avg} = \frac{0}{10 \text{ s}} = 0 \text{ m/s}$ .

Why other options are wrong:
- Option A: $2 \text{ ms}^{-1}$ : This is incorrect because it likely calculates average speed instead of average velocity.
- Option B: $3.14 \text{ ms}^{-1}$ : This is incorrect because it likely calculates average speed based on half the circumference.
- Option C: $6.28 \text{ ms}^{-1}$ : This is incorrect because it likely calculates average speed based on the full circumference.

Key Concepts:
- Average velocity
- Displacement
- Circular motion

Memory Aids:
- Remember that displacement is the change in position, not the total distance traveled.

Quick Tips:
- When dealing with circular motion, remember that the displacement is zero for a complete revolution.

4 Easy

4 of 31

Application +4 | -1

The motion of a particle along a straight line is described by equation

x=2+12 t-t^2

, where

x

is in metre and

t

is in second. The retardation of the particle when its velocity becomes zero is

View answer and explanation

Correct option: A

2 \mathrm{~m} / \mathrm{s}^2

Option A (

2 \mathrm{~m} / \mathrm{s}^2

) is correct because the retardation is the magnitude of the acceleration when the velocity is zero.

Solution approach

Correct Answer + Core Reason: $2 \mathrm{~m} / \mathrm{s}^2$ . The retardation is the magnitude of the acceleration when velocity is zero.

Step-by-Step Solution:

Step 1: Given the equation of motion: $x=2+12 t-t^2$ .

Step 2: Find the velocity by differentiating the position equation with respect to time: $v = \frac{dx}{dt} = 12 - 2t$ .

Step 3: Find the time when the velocity is zero: $v = 0 \Rightarrow 12 - 2t = 0 \Rightarrow t = 6 \mathrm{~s}$ .

Step 4: Find the acceleration by differentiating the velocity equation with respect to time: $a = \frac{dv}{dt} = -2 \mathrm{~m} / \mathrm{s}^2$ .

Step 5: Find the retardation, which is the magnitude of the acceleration: Retardation $= |a| = |-2| = 2 \mathrm{~m} / \mathrm{s}^2$ .

Why other options are wrong:
- **Option B ( $12 \mathrm{~m} / \mathrm{s}^2$ ): This is incorrect because it is not the retardation. It is related to the initial velocity, not the acceleration.
- Option C ( $-2 \mathrm{~m} / \mathrm{s}^2$ ): This is incorrect because retardation is the magnitude of acceleration, so it cannot be negative.
- Option D ( $-12 \mathrm{~m} / \mathrm{s}^2$ ): This is incorrect because it is not the retardation and it has the wrong sign.

Key Concepts:
- Kinematics
- Velocity as the derivative of position
- Acceleration as the derivative of velocity
- Retardation as the magnitude of negative acceleration

Memory Aids:
- Remember that velocity is the rate of change of position, and acceleration is the rate of change of velocity.

Quick Tips:**
- Always differentiate the position equation twice to find the acceleration.
- Retardation is always a positive value.

5 Easy

5 of 31

Understanding +4 | -1

Which of the following statements is true for a car moving on the road?

View answer and explanation

Correct option: B

With respect to the frame of reference attached to the person sitting in the car, the car is at rest.

Option B (With respect to the frame of reference attached to the person sitting in the car, the car is at rest) is correct because the car's position doesn't change relative to the person inside.

Solution approach

Correct Answer + Core Reason: With respect to the frame of reference attached to the person sitting in the car, the car is at rest. Relative to the person inside, the car's position doesn't change.

Step-by-Step Solution:

Step 1: Define a frame of reference: A frame of reference is a coordinate system used to describe the motion of an object.

Step 2: Consider the frame of reference attached to the ground: From this perspective, the car is moving.

Step 3: Consider the frame of reference attached to a person sitting in the car: From this perspective, the car is at rest because the person and the car are moving together.

Step 4: Consider the frame of reference attached to a person outside the car: From this perspective, the car is moving.

Why other options are wrong:
- Option A: With respect to the frame of reference attached to the ground, the car is at rest. - This is incorrect because the car is moving relative to the ground.
- Option C: With respect to the frame of reference attached to the person outside the car, the car is at rest. - This is incorrect because the car is moving relative to a person outside the car.
- Option D: None of the above - Option B is correct.

Key Concepts:
- Frame of Reference
- Relative Motion

Memory Aids:
- Think about your own experience in a moving car.

Quick Tips:
- Always consider the observer's perspective when analyzing motion.

6 Easy

6 of 31

Understanding +4 | -1

A body thrown vertically upward with an initial velocity of 1.4 m/s returns in 2s. The total displacement of the ball is

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because the total displacement of the ball is zero since it returns to its starting point.

Solution approach

Correct Answer + Core Reason: Zero. Displacement is the shortest distance between initial and final points. Since the ball returns to the starting point, the displacement is zero.

Step-by-Step Solution:

Step 1: Define displacement: Displacement is the change in position of an object.

Step 2: Identify initial and final positions: The ball starts at a point and returns to the same point.

Step 3: Calculate displacement: Since the initial and final positions are the same, the displacement is zero.

Why other options are wrong:
- Option A: This is incorrect because displacement is zero when the object returns to its starting point.
- Option B: This is incorrect because displacement is zero when the object returns to its starting point.
- Option C: This is incorrect because displacement is zero when the object returns to its starting point.

Key Concepts:
- Displacement
- Initial and final positions

Memory Aids:
- Displacement is a vector quantity; distance is a scalar quantity.

Quick Tips:
- Remember that displacement can be zero even if the distance traveled is not zero.

7 Easy

7 of 31

Application +4 | -1

If a cyclist takes one minute to complete half revolution on a circular path 120 m radius. What is the average velocity?

View answer and explanation

Correct option: D

4 m/s

Option D (4 m/s) is correct because average velocity is calculated by dividing the displacement by the time taken.

Solution approach

Correct Answer + Core Reason: 4 m/s. Average velocity is calculated by dividing the displacement by the time taken.

Step-by-Step Solution:

Step 1: Identify the given values: Radius $R = 120 \text{{ m}}$ , Time $t = 1 \text{{ minute}} = 60 \text{{ s}}$

Step 2: Calculate the displacement: Displacement = $2R = 2 \times 120 = 240 \text{{ m}}$

Step 3: Calculate the average velocity: $V_{avg} = \frac{{\text{{Displacement}}}}{{\text{{Time}}}} = \frac{{240}}{{60}}$

Step 4: Simplify: $V_{avg} = 4 \text{{ m/s}}$

Step 5: Final Answer: $4 \text{{ m/s}}$

Why other options are wrong:
- Option A (1 m/s): This is incorrect because it's likely a calculation error in applying the formula.
- Option B (2 m/s): This is incorrect because it's likely a calculation error in applying the formula.
- Option C (3 m/s): This is incorrect because it's likely a calculation error in applying the formula.

Key Concepts:
- Average velocity
- Displacement

Memory Aids:
- Remember the formula: $V_{avg} = \frac{{\text{{Displacement}}}}{{\text{{Time}}}}$

Quick Tips:
- Ensure the units are consistent before substituting values.
- Pay attention to the given values and the formula.

8 Easy

8 of 31

Understanding +4 | -1

Assertion: The actual distance covered by an object in a given time interval can be equal to or greater than the magnitude of the displacement.

Reason: The distance covered is a scalar quantity, while the displacement is a vector quantity.

View answer and explanation

Correct option: B

If both assertion and reason are true, but reason is not correct explanation of the assertion.

Option B (If both assertion and reason are true, but reason is not correct explanation of the assertion) is correct because both statements are true, but the scalar/vector nature doesn't directly explain the relationship between distance and displacement.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true, but reason is not correct explanation of the assertion. Distance can be greater than or equal to displacement, and distance is a scalar while displacement is a vector.

Step-by-Step Solution:

Step 1: Understand the assertion: The actual distance covered can be equal to or greater than the magnitude of the displacement.

Step 2: Understand the reason: Distance is a scalar quantity, while displacement is a vector quantity.

Step 3: The assertion is true because distance is the total path length, while displacement is the shortest distance between the initial and final points.

Step 4: The reason is also true, as distance is a scalar and displacement is a vector.

Step 5: However, the reason doesn't directly explain why distance can be greater than displacement. The explanation lies in the path taken by the object.

Why other options are wrong:
- Option A: The reason is true, but it doesn't explain the assertion.
- Option C: Both assertion and reason are true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Distance
- Displacement
- Scalar Quantity
- Vector Quantity

Memory Aids:
- Distance: total path length
- Displacement: shortest distance

Quick Tips:
- Remember the difference between distance and displacement.

9 Easy

9 of 31

Application +4 | -1

A police jeep is chasing with velocity of

45 \mathrm{~km} / \mathrm{h}

a thief in another jeep moving with velocity

153 \mathrm{~km} / \mathrm{h}

. Police fires a bullet with muzzle velocity of

180 \mathrm{~m} / \mathrm{~s}

. The velocity with which it will strike the car of the thief is

View answer and explanation

Correct option: A

150 \mathrm{~m} / \mathrm{s}

Option A (

150 \mathrm{~m} / \mathrm{s}

) is correct because it is the relative velocity of the bullet with respect to the thief's car.

Solution approach

Correct Answer + Core Reason: $150 \mathrm{~m} / \mathrm{s}$ . The relative velocity is calculated by subtracting the thief's velocity from the bullet's effective velocity.

Step-by-Step Solution:

Step 1: Convert the velocities of the police jeep and the thief's jeep from km/h to m/s.\nPolice jeep: $v_p = 45 \frac{{\text{{km}}}}{{\text{{h}}}} = 45 \times \frac{{1000 \text{{ m}}}}{{3600 \text{{ s}}}} = 12.5 \text{{ m/s}}$ .\nThief's jeep: $v_t = 153 \frac{{\text{{km}}}}{{\text{{h}}}} = 153 \times \frac{{1000 \text{{ m}}}}{{3600 \text{{ s}}}} = 42.5 \text{{ m/s}}$ .

Step 2: Calculate the effective velocity of the bullet with respect to the ground: $v_b = \text{{muzzle velocity}} + v_p = 180 + 12.5 = 192.5 \text{{ m/s}}$ .

Step 3: Calculate the relative velocity of the bullet with respect to the thief's car: $v_{rel} = v_b - v_t = 192.5 - 42.5 = 150 \text{{ m/s}}$ .

Final Answer: $150 \mathrm{~m} / \mathrm{s}$

Why other options are wrong:
- **Option B ( $27 \mathrm{~m} / \mathrm{s}$ ): This is incorrect; it's likely a result of subtracting the velocities without considering the bullet's muzzle velocity.
- Option C ( $450 \mathrm{~m} / \mathrm{s}$ ): This is incorrect; it's a significantly larger value and doesn't represent the relative velocity.
- Option D ( $250 \mathrm{~m} / \mathrm{s}$ ): This is incorrect; it's a larger value than the correct relative velocity.

Key Concepts:
- Relative velocity
- Velocity addition
- Unit conversion

Memory Aids:
- Remember to convert units to be consistent (km/h to m/s).
- The relative velocity is the difference between the two velocities.

Quick Tips:**
- Always convert units to the same system before performing calculations.
- Pay attention to the frame of reference when calculating relative velocities.

10 Easy

10 of 31

Analyzing +4 | -1

A particle is moving on a straight line path with constant acceleration directed along the direction of instantaneous velocity. Which of the following statements are false about the motion of particle?

View answer and explanation

Correct option: D

All of the above

Option D is correct. If the acceleration is in the direction of the instantaneous velocity, the particle will not reverse direction, and the distance covered will equal the magnitude of the displacement. The magnitude of the average velocity will equal the average speed.

Solution approach

Correct Answer + Core Reason: All of the above. If acceleration is along the velocity, the particle cannot reverse direction, distance equals displacement magnitude, and average velocity magnitude equals average speed.

Step-by-Step Solution:

Step 1: Analyze the given condition: The particle is moving on a straight line with constant acceleration directed along the direction of instantaneous velocity.

Step 2: If the acceleration is in the same direction as the velocity, the particle will continuously increase its speed in that direction. It cannot reverse its direction.

Step 3: Since the particle is not reversing direction, the distance covered is equal to the magnitude of the displacement.

Step 4: Since the distance covered is equal to the magnitude of the displacement, the magnitude of average velocity is equal to the average speed.

Step 5: Therefore, all the given statements are false.

Why other options are wrong:
- Option A: The particle cannot reverse direction.
- Option B: Distance covered is equal to the magnitude of displacement.
- Option C: The magnitude of average velocity is equal to the average speed.

Key Concepts:
- Kinematics
- Acceleration
- Velocity
- Displacement
- Distance

Memory Aids:
- Remember the definitions of displacement, distance, velocity, and speed.

Quick Tips:
- Always consider the direction of acceleration and velocity.

11 Easy

11 of 31

Analyzing +4 | -1

The velocity

\mathrm{v}

of a particle is given by the equation

v=6 t^2-6 t^3

where

v

is in

\mathrm{m} / \mathrm{sec}

and

\mathrm{t}

is time in seconds then :

View answer and explanation

Correct option: C

minimum velocity is zero

Option D (minimum velocity is

-2 \mathrm{~m} / \mathrm{sec}

) is incorrect. The minimum velocity is zero, occurring at t=0.

Solution approach

Correct Answer + Core Reason: minimum velocity is zero. The minimum velocity occurs at t=0 and is equal to 0 m/s.

Step-by-Step Solution:

Step 1: Given velocity equation: $v = 6t^2 - 6t^3$

Step 2: To find the times at which velocity is maximum or minimum, differentiate $v$ with respect to $t$ and set it to zero:
$\frac{dv}{dt} = 12t - 18t^2 = 0$
$6t(2 - 3t) = 0$
$t = 0$ or $t = \frac{2}{3}$

Step 3: To determine whether these times correspond to a maximum or minimum, find the second derivative:
$\frac{d^2v}{dt^2} = 12 - 36t$

Step 4: Evaluate the second derivative at $t = 0$ :
$\frac{d^2v}{dt^2}|_{t=0} = 12 > 0$ , which indicates a minimum.\nEvaluate the second derivative at $t = \frac{2}{3}$ :
$\frac{d^2v}{dt^2}|_{t=2/3} = 12 - 36(\frac{2}{3}) = 12 - 24 = -12 < 0$ , which indicates a maximum.

Step 5: Calculate the velocity at $t = 0$ :\nv(0) = 6(0)² - 6(0)³ = 0 m/s

Step 6: Calculate the velocity at $t = \frac{2}{3}$ :\nv( $\frac{2}{3}$ ) = 6( $\frac{2}{3}$ )² - 6( $\frac{2}{3}$ )³ = 6( $\frac{4}{9}$ ) - 6( $\frac{8}{27}$ ) = $\frac{24}{9}$ - $\frac{48}{27}$ = $\frac{72 - 48}{27}$ = $\frac{24}{27}$ = $\frac{8}{9}$ m/s

Step 7: The minimum velocity is 0 m/s.

Why other options are wrong:
- Option A (at t=0, velocity is maximum): The velocity is minimum at t=0.
- Option B (at t=2/3, velocity is minimum): The velocity is maximum at t=2/3.
- Option D (minimum velocity is -2 m/sec): The minimum velocity is 0 m/s.

Key Concepts:
- Differentiation
- Finding maxima and minima
- Velocity as a function of time

Memory Aids:
- Remember that the first derivative gives the slope, and the second derivative indicates concavity.

Quick Tips:
- Always check the second derivative to confirm whether a critical point is a maximum or minimum.

12 Easy

12 of 31

Understanding +4 | -1

If an object is falling freely in vacuum, then body moves with

View answer and explanation

Correct option: B

Uniform acceleration

Option B (Uniform acceleration) is correct because an object falling freely in a vacuum experiences constant acceleration due to gravity.

Solution approach

Correct Answer + Core Reason: Uniform acceleration. An object falling freely in a vacuum experiences constant acceleration due to gravity.

Step-by-Step Solution:

Step 1: Recall that free fall means the only force acting on the object is gravity.

Step 2: In a vacuum, there is no air resistance.

Step 3: Therefore, the net force on the object is constant and equal to its weight, $F = mg$ , where $m$ is the mass and $g$ is the acceleration due to gravity.

Step 4: According to Newton's second law, $F = ma$ , so $mg = ma$ , which means $a = g$ .

Step 5: Since $g$ is constant, the acceleration is uniform.

Step 6: Final Answer: Uniform acceleration.

Why other options are wrong:
- Option A: The velocity increases linearly with time due to the constant acceleration, so it's not uniform.
- Option C: The acceleration is constant, not variable.
- Option D: The speed increases linearly with time due to the constant acceleration, so it's not uniform.

Key Concepts:
- Free fall
- Acceleration due to gravity
- Newton's second law

Memory Aids:
- Free fall = constant acceleration.

Quick Tips:
- In free fall problems, assume constant acceleration due to gravity unless otherwise stated.

13 Easy

13 of 31

Understanding +4 | -1

The velocity-time graph of particle comes out to be a non-linear curve. The motion is

View answer and explanation

Correct option: C

non-uniform accelerated motion

Option C (non-uniform accelerated motion) is correct because a non-linear velocity-time graph indicates that the acceleration is changing with time.

Solution approach

Correct Answer + Core Reason: non-uniform accelerated motion. A non-linear velocity-time graph indicates that the acceleration is not constant.

Step-by-Step Solution:
Step 1: The slope of a velocity-time graph represents the acceleration of the particle.
Step 2: If the velocity-time graph is a straight line, the slope is constant, indicating uniform acceleration.
Step 3: If the velocity-time graph is a non-linear curve, the slope is not constant, indicating non-uniform acceleration.
Step 4: Therefore, a non-linear velocity-time graph represents non-uniform accelerated motion.

Why other options are wrong:
- Option A: uniform velocity motion: This would be represented by a horizontal line on the velocity-time graph.
- Option B: uniformly accelerated motion: This would be represented by a straight line on the velocity-time graph.
- Option D: Nothing can be said about the motion: The velocity-time graph provides information about the acceleration.

Key Concepts:
- Velocity-time graph
- Acceleration
- Uniform and non-uniform motion

Memory Aids:
- Slope of v-t graph is acceleration.

Quick Tips:
- Remember that the shape of the velocity-time graph provides information about the type of motion.

14 Easy

14 of 31

Analyzing +4 | -1

Statement-1: Two balls of different masses are thrown vertically upwards with same speed. They will pass through their point of projection in the downward direction with the same speed in absence of air resistance.
Statement-2: In absence of air resistance, the mechanical energy of a projectile is conserved.

View answer and explanation

Correct option: A

Statement-1 and Statement-2 are true, Statement-2 is the correct explanation for Statement-1

Statement-1 and Statement-2 are true, and Statement-2 is the correct explanation for Statement-1 because the conservation of mechanical energy implies the same kinetic energy (and thus speed) at the same height.

Solution approach

Correct Answer + Core Reason: Statement-1 and Statement-2 are true, Statement-2 is the correct explanation for Statement-1 because mechanical energy conservation implies the same speed at the same height.

Step-by-Step Solution:

Step 1: Define the initial conditions: Two balls of different masses are thrown vertically upwards with the same initial speed.

Step 2: Consider the conservation of mechanical energy. In the absence of air resistance, the total mechanical energy (potential + kinetic) remains constant.

Step 3: At the point of projection, the potential energy is zero, and the kinetic energy is $\frac{1}{2}mv^2$ . The total mechanical energy is $E = \frac{1}{2}mv^2$ .

Step 4: When the ball returns to the point of projection, its height is again zero, so the potential energy is zero. The kinetic energy must be equal to the initial kinetic energy due to conservation of energy.

Step 5: Therefore, $\frac{1}{2}mv_f^2 = \frac{1}{2}mv^2$ , which implies $v_f = v$ . The final speed is the same as the initial speed, regardless of the mass.

Step 6: Since mechanical energy is conserved, the balls will have the same kinetic energy at the same height and thus will have the same speed.

Why other options are wrong:
- Option B: This option is incorrect because Statement-2 *is* the correct explanation for Statement-1.
- Option C: This option is incorrect because Statement-2 is true.
- Option D: This option is incorrect because Statement-1 is true.

Key Concepts:
- Conservation of mechanical energy
- Kinetic energy
- Potential energy

Memory Aids:
- Remember that in the absence of non-conservative forces, mechanical energy is conserved.

Quick Tips:
- When dealing with projectile motion without air resistance, consider using conservation of energy.

15 Easy

15 of 31

Understanding +4 | -1

Velocity of a body on reaching the point from which it was projected upwards, is

View answer and explanation

Correct option: D

v = - u

Option D (v = - u) is correct because the velocity of a body on reaching the point of projection is equal in magnitude but opposite in direction to the initial velocity.

Solution approach

Correct Answer + Core Reason: v = - u. The final velocity is equal in magnitude but opposite in direction to the initial velocity.

Step-by-Step Solution:

Step 1: Consider a body projected upwards with initial velocity 'u'.

Step 2: Due to gravity, the body decelerates and eventually reaches its highest point where its velocity is zero.

Step 3: As it falls back down, it accelerates due to gravity.

Step 4: When it reaches the point of projection, its speed will be equal to the initial speed 'u', but its direction will be downwards.

Step 5: Therefore, the final velocity v = -u.

Why other options are wrong:
- Option A: This is incorrect; the velocity is not zero when it reaches the point of projection.
- Option B: This is incorrect; the velocity is not twice the initial velocity.
- Option C: This is incorrect; the velocity is not half the initial velocity.

Key Concepts:
- Projectile Motion
- Kinematics

Memory Aids:
- Final velocity = - Initial velocity (when returning to the same height).

Quick Tips:
- Remember that velocity is a vector quantity.

16 Easy

16 of 31

Application +4 | -1

A cat wants to catch a rat. The cat follows the path whose equation is

x+y=0

, but the rat follows the path whose equation is

x^2+y^2=4

. The coordinates of possible points of catching the rat are

View answer and explanation

Correct option: B

(-\sqrt{2}, \sqrt{2})

Option B

((-\sqrt{2}, \sqrt{2}))

is correct because it satisfies both the equation of the cat's path (

x+y=0

) and the rat's path (

x^2+y^2=4

).

Solution approach

Correct Answer + Core Reason: $(-\sqrt{2}, \sqrt{2})$ . This point satisfies both equations, representing the intersection of the cat's and rat's paths.

Step-by-Step Solution:

Step 1: The cat's path is given by $x + y = 0$ , which implies $y = -x$ .

Step 2: The rat's path is given by $x^2 + y^2 = 4$ .

Step 3: Substitute $y = -x$ into the equation of the rat's path: $x^2 + (-x)^2 = 4$ .

Step 4: Simplify the equation: $2x^2 = 4$ , which gives $x^2 = 2$ .

Step 5: Solve for $x$ : $x = \pm \sqrt{2}$ .

Step 6: Find the corresponding values of $y$ : If $x = \sqrt{2}$ , then $y = -\sqrt{2}$ . If $x = -\sqrt{2}$ , then $y = \sqrt{2}$ .

Step 7: The possible points of catching the rat are $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$ .

Step 8: Check the given options. Only $(-\sqrt{2}, \sqrt{2})$ is among the options.

Why other options are wrong:
- **Option A $((\sqrt{2}, \sqrt{2}))$ :** This point does not satisfy the equation $x + y = 0$ because $\sqrt{2} + \sqrt{2} \neq 0$ .
- **Option C $((\sqrt{2}, \sqrt{3}))$ : This point does not satisfy either equation.
- Option D $((0,0))$ :** This point satisfies $x + y = 0$ but does not satisfy $x^2 + y^2 = 4$ because $0^2 + 0^2 \neq 4$ .

Key Concepts:
- Solving simultaneous equations
- Equation of a line
- Equation of a circle

Memory Aids:
- Remember the equation of a circle: $x^2 + y^2 = r^2$

Quick Tips:
- Always check if the solution satisfies both equations.

17 Easy

17 of 31

Analyzing +4 | -1

The distance travelled by a particle is directly proportional to

\sqrt{t}

, where

t

is time elapsed. Which of the following is correct regarding the motion of the particle?

View answer and explanation

Correct option: C

Particle is moving with decreasing retardation

Option C (Particle is moving with decreasing retardation) is correct because the acceleration is negative and its magnitude decreases with time, indicating decreasing retardation.

Solution approach

Correct Answer + Core Reason: Particle is moving with decreasing retardation. The acceleration is negative, and its magnitude decreases with time, indicating decreasing retardation.

Step-by-Step Solution:

Step 1: Express the distance traveled as a function of time: $x = k\sqrt{t}$ , where $k$ is a constant.

Step 2: Find the velocity by differentiating the distance with respect to time: $v = \frac{dx}{dt} = \frac{k}{2}t^{-1/2}$ .

Step 3: Find the acceleration by differentiating the velocity with respect to time: $a = \frac{dv}{dt} = -\frac{k}{4}t^{-3/2}$ .

Step 4: Analyze the acceleration. The acceleration is negative, indicating retardation. The magnitude of the acceleration is $|a| = \frac{k}{4}t^{-3/2} = \frac{k}{4t^{3/2}}$ . As time $t$ increases, the magnitude of the acceleration decreases. Therefore, the particle is moving with decreasing retardation.

Why other options are wrong:
- Option A: The acceleration is decreasing, not increasing.
- Option B: The acceleration is decreasing, but it's retardation.
- Option D: The retardation is decreasing, not increasing.

Key Concepts:
- Kinematics
- Velocity
- Acceleration
- Retardation

Memory Aids:
- Remember that retardation is negative acceleration.

Quick Tips:
- Differentiate the distance function to find velocity and acceleration.

18 Easy

18 of 31

Application +4 | -1

Two stones of masses

2 \mathrm{~kg}

and

4 \mathrm{~kg}

are dropped from height

30 \mathrm{~m}

and

40 \mathrm{~m}

respectively. The ratio of time taken by the stones to reach the ground is

View answer and explanation

Correct option: D

\sqrt{\frac{3}{4}}

Option D (

\sqrt{\frac{3}{4}}

) is correct because the time taken to reach the ground is proportional to

\sqrt{h}

, so the ratio of times is

\sqrt{\frac{30}{40}} = \sqrt{\frac{3}{4}}

.

Solution approach

Correct Answer + Core Reason: $\sqrt{\frac{3}{4}}$ . The ratio of time taken is proportional to the square root of the ratio of heights.

Step-by-Step Solution:

Step 1: The time taken for an object to fall from a height $h$ is given by $t = \sqrt{\frac{{2h}}{g}}$ , where $g$ is the acceleration due to gravity.

Step 2: For the first stone, $h_1 = 30 \text{ m}$ , so $t_1 = \sqrt{\frac{{2h_1}}{g}} = \sqrt{\frac{{2(30)}}{g}}$ .

Step 3: For the second stone, $h_2 = 40 \text{ m}$ , so $t_2 = \sqrt{\frac{{2h_2}}{g}} = \sqrt{\frac{{2(40)}}{g}}$ .

Step 4: The ratio of the times is $\frac{{t_1}}{{t_2}} = \frac{{\sqrt{\frac{{2(30)}}{g}}}}{{\sqrt{\frac{{2(40)}}{g}}}} = \sqrt{\frac{{30}}{{40}}} = \sqrt{\frac{3}{4}}$ .

Why other options are wrong:
- **Option A ( $\frac{1}{2}$ ): Incorrect calculation of the ratio of times.
- Option B ( $1: 1$ ): Incorrect calculation of the ratio of times.
- Option C ( $\frac{4}{{\sqrt{3}}}}$ ): Incorrect calculation of the ratio of times.

Key Concepts:
- Free fall
- Kinematics

Memory Aids:**
- Remember the formula for time of free fall: $t = \sqrt{\frac{{2h}}{g}}$ .

Quick Tips:
- The mass of the stones does not affect the time taken to reach the ground in this case (neglecting air resistance).
- The ratio of times is proportional to the square root of the ratio of heights.

19 Easy

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Application +4 | -1

A person walks up a stalled escalator in

90 \mathrm{~s}

. When he is just standing on the same moving escalator, then he is carried for

60 \mathrm{~s}

. The time it would take him to walk up the moving escalator will be

View answer and explanation

Correct option: D

36 \mathrm{~s}

Option D (

36 \mathrm{~s}

) is correct because the relative speeds add up when the person walks on the moving escalator, and the time taken is calculated using the combined speed.

Solution approach

Correct Answer + Core Reason: $36 \mathrm{~s}$ . The time is calculated using the combined speed of the person and the escalator.

Step-by-Step Solution:

Step 1: Define the distance to be covered as $s$ .

Step 2: Define the person's speed as $v_1 = \frac{s}{90}$ (when the escalator is stalled).

Step 3: Define the escalator's speed as $v_2 = \frac{s}{60}$ (when the person is standing still).

Step 4: When the person walks up the moving escalator, their speeds add up: $v = v_1 + v_2$ .

Step 5: Calculate the combined speed: $v = \frac{s}{90} + \frac{s}{60} = s(\frac{1}{90} + \frac{1}{60}) = s(\frac{2+3}{180}) = \frac{5s}{180} = \frac{s}{36}$ .

Step 6: Calculate the time it takes to walk up the moving escalator: $t = \frac{s}{v} = \frac{s}{\frac{s}{36}} = 36 \mathrm{~s}$ .

Why other options are wrong:
- **Option A ( $27 \mathrm{~s}$ ): This is incorrect and likely arises from an error in calculating the combined speed or misunderstanding the relationship between speed, distance, and time.
- Option B ( $50 \mathrm{~s}$ ): This is incorrect and suggests an error in calculating the combined speed, possibly by subtracting the speeds instead of adding them.
- Option C ( $18 \mathrm{~s}$ ): This is incorrect and likely arises from an error in calculating the combined speed, possibly by using a wrong formula.

Key Concepts:
- Relative Motion
- Speed, Distance, and Time

Memory Aids:**
- $v = \frac{s}{t}$ (Speed = Distance / Time)

Quick Tips:
- When objects move in the same direction, their speeds add up.
- Be careful with fractions when calculating combined speeds.

20 Easy

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Analyzing +4 | -1

A particle is moving along

x

-axis whose acceleration is given by

a=3 x-4

, where

x

is the location of the particle. At

t=0

, the particle is at rest at

x=4 / 3

. The distance travelled by the particle in 5 seconds is

View answer and explanation

Correct option: A

zero

Option A (zero) is correct because the particle starts at rest with zero acceleration, so it remains at rest and travels no distance.

Solution approach

Correct Answer + Core Reason: zero. Since the particle starts at rest with zero acceleration, it remains at rest.

Step-by-Step Solution:

Step 1: Given $a = 3x - 4$ , where a is the acceleration and x is the position of the particle.

Step 2: At $t = 0$ , the particle is at rest, so $v = 0$ , and its position is $x = \frac{4}{3}$ .

Step 3: Calculate the acceleration at $t = 0$ :
$a = 3(\frac{4}{3}) - 4 = 4 - 4 = 0$

Step 4: Since the particle is at rest ( $v = 0$ ) and has zero acceleration ( $a = 0$ ) at $t = 0$ , it will remain at rest for all time.

Step 5: Therefore, the distance traveled by the particle in 5 seconds is zero.

Why other options are wrong:
- Option B: $\approx 42 \mathrm{~m}$ - This is incorrect because the particle does not move.
- Option C: infinite - This is incorrect because the particle does not move.
- Option D: None of these - This is incorrect because the distance traveled is zero.

Key Concepts:
- Kinematics
- Acceleration
- Velocity
- Rest

Memory Aids:
- If an object is at rest and has zero acceleration, it will remain at rest.

Quick Tips:
- Always check the initial conditions (position, velocity, acceleration) to determine the motion of the particle.

21 Easy

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Application +4 | -1

An automobile travels on a straight road for

40 \mathrm{~km}

at

30 \mathrm{~km}

\mathrm{h}^{-1}

. It then continues in the same direction for another

40 \mathrm{km}

at

60 \mathrm{~km} \mathrm{~h}^{-1}

. What is the average velocity of the car during its

80 \mathrm{~km}

trip?

View answer and explanation

Correct option: C

40 \mathrm{~km} \mathrm{~h}^{-1}

Option C (40 km/h) is correct because the average velocity is calculated by dividing the total displacement by the total time taken for the entire trip.

Solution approach

Correct Answer + Core Reason: $40 \mathrm{~km} \mathrm{~h}^{-1}$ . The average velocity is calculated by dividing the total distance by the total time.

Step-by-Step Solution:

Step 1: Calculate the time taken for the first part of the journey: $\Delta t_1 = \frac{{40}}{{30}} = 1.33 \mathrm{~h}$ .

Step 2: Calculate the time taken for the second part of the journey: $\Delta t_2 = \frac{{40}}{{60}} = 0.67 \mathrm{~h}$ .

Step 3: Calculate the total displacement: $\Delta x = 40 + 40 = 80 \mathrm{~km}$ .

Step 4: Calculate the total time: $\Delta t = \Delta t_1 + \Delta t_2 = 1.33 + 0.67 = 2 \mathrm{~h}$ .

Step 5: Calculate the average velocity: $v_{\text {av }} = \frac{{\Delta x}}{{\Delta t}} = \frac{{80}}{2} = 40 \mathrm{~km} \mathrm{~h}^{-1}$ .

Step 6: Final Answer: $40 \mathrm{~km} \mathrm{~h}^{-1}$ .

Why other options are wrong:
- Option A: $30 \mathrm{~km} \mathrm{~h}^{-1}$ This is incorrect because it only considers the velocity of the first part of the journey.
- Option B: $50 \mathrm{~km} \mathrm{~h}^{-1}$ This is incorrect because it is a simple average of the two velocities, which is not the correct way to calculate average velocity when the distances are the same but the times are different.
- Option D: $60 \mathrm{~km} \mathrm{~h}^{-1}$ This is incorrect because it only considers the velocity of the second part of the journey.

Key Concepts:
- Average velocity
- Displacement
- Time interval

Memory Aids:
- Remember the formula for average velocity: $v_{\text {av }} = \frac{{\Delta x}}{{\Delta t}}$ .

Quick Tips:
- Always calculate the total displacement and total time before calculating the average velocity.
- Be careful not to simply average the velocities if the time intervals or distances are different.

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Understanding +4 | -1

A ball of mass

{m_1}

and another ball of mass

{m_2}

are dropped from equal height. If time taken by the balls are

{t_1}

and

{t_2}

respectively, then

View answer and explanation

Correct option: B

{t_1} = {t_2}

Option B (

{t_1} = {t_2}

) is correct because the time taken for an object to fall from a certain height is independent of its mass, assuming no air resistance.

Solution approach

Correct Answer + Core Reason: ${t_1} = {t_2}$ . The time of fall is independent of mass.

Step-by-Step Solution:

Step 1: Recall the equation for the time of fall under gravity: $h = \frac{1}{2} g t^2$ , where h is the height, g is the acceleration due to gravity, and t is the time.

Step 2: Solve for time: $t = \sqrt{\frac{2h}{g}}$ .

Step 3: Analyze the equation: The time depends only on the height (h) and the acceleration due to gravity (g), not on the mass of the ball.

Step 4: Since both balls are dropped from the same height, the time taken will be the same: $t_1 = t_2$ .

Why other options are wrong:
- Option A: The time of fall is independent of mass.
- Option C: The time of fall is independent of mass.
- Option D: The time of fall is independent of mass.

Key Concepts:
- Free Fall
- Acceleration due to Gravity
- Kinematics

Memory Aids:
- Remember that in the absence of air resistance, all objects fall with the same acceleration regardless of their mass.

Quick Tips:
- The time of fall depends only on height and gravity.

23 Easy

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Understanding +4 | -1

Let

\vec{v}

and

\vec{a}

denote the velocity and acceleration respectively of a body in one dimensional motion. Then correct option is

View answer and explanation

Correct option: C

Speed will decrease when

\vec{V}

and

\vec{a}

anti-parallel

Option C is correct because speed decreases when velocity and acceleration are anti-parallel, meaning they are in opposite directions.

Solution approach

Correct Answer + Core Reason: Speed will decrease when $\vec{V}$ and $\vec{a}$ anti-parallel. When velocity and acceleration are in opposite directions, the object slows down.

Step-by-Step Solution:

Step 1: Define velocity and acceleration: Velocity ( $\vec{v}$ ) is the rate of change of displacement, and acceleration ( $\vec{a}$ ) is the rate of change of velocity.

Step 2: Consider anti-parallel vectors: If $\vec{v}$ and $\vec{a}$ are anti-parallel, it means they point in opposite directions.

Step 3: Effect on speed: When acceleration is in the opposite direction to velocity, the object slows down, i.e., its speed decreases.

Why other options are wrong:
- Option A: If $\vec{a} = 0$ , the velocity is constant, so the magnitude of velocity (speed) does not necessarily decrease.
- Option B: Speed increases when acceleration is in the same direction as velocity. If acceleration is opposite to velocity, speed decreases.
- Option D: Options A and B are incorrect, so this option is also incorrect.

Key Concepts:
- Velocity
- Acceleration
- Parallel and anti-parallel vectors

Memory Aids:
- Acceleration in the direction of velocity increases speed; acceleration opposite to velocity decreases speed.

Quick Tips:
- Visualize the motion: If an object is moving to the right and accelerating to the left, it will slow down.

24 Easy

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Application +4 | -1

Two particles are moving with velocities 17 m/sec and 14 m/sec along positive X-axis. Their initial positions are at x = 24 m, x = 120 m respectively. find time taken by first particle to cross second one

View answer and explanation

Correct option: B

32 sec

Option B (32 sec) is correct because it is the time when both particles have the same position, calculated by equating their position equations.

Solution approach

Correct Answer + Core Reason: 32 sec. The time is calculated by equating the positions of the two particles.

Step-by-Step Solution:

Step 1: Identify the given values: $v_1 = 17$ m/s, $v_2 = 14$ m/s, $x_1 = 24$ m, $x_2 = 120$ m.

Step 2: Write the position equations for both particles: $x_1(t) = x_1 + v_1t$ and $x_2(t) = x_2 + v_2t$ .

Step 3: Set the positions equal to each other: $x_1 + v_1t = x_2 + v_2t$ .

Step 4: Rearrange the equation to solve for t: $v_1t - v_2t = x_2 - x_1 \Rightarrow t(v_1 - v_2) = x_2 - x_1 \Rightarrow t = \frac{{x_2 - x_1}}{{v_1 - v_2}}$ .

Step 5: Substitute the given values: $t = \frac{{120 - 24}}{{17 - 14}} = \frac{{96}}{3}$ .

Step 6: Calculate the time: $t = 32$ s.

Why other options are wrong:
- Option A (50 sec): This is incorrect and likely results from an error in the calculation or an incorrect setup of the equation.
- Option C (30 sec): This is incorrect and likely results from an error in the calculation or an incorrect setup of the equation.
- Option D (45 sec): This is incorrect and likely results from an error in the calculation or an incorrect setup of the equation.

Key Concepts:
- Kinematics
- Uniform motion
- Position as a function of time

Memory Aids:
- Remember the equation for position with constant velocity: $x(t) = x_0 + vt$

Quick Tips:
- Draw a diagram to visualize the problem.
- Write down the given values and the equations of motion for each object.
- Set the positions equal to each other to find the time when they meet.

25 Easy

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Application +4 | -1

The displacement of particle is given by

y = 2 + 3t + 8{t^2} - 6{t^4}\,\,{\text{m}}

then initial velocity is given by (m/s)

View answer and explanation

Correct option: C

3

Option C (3) is correct because the initial velocity is the derivative of the displacement equation evaluated at t=0.

Solution approach

Correct Answer + Core Reason: 3. The initial velocity is the derivative of the displacement equation evaluated at t=0.

Step-by-Step Solution:

Step 1: Given the displacement equation: $y = 2 + 3t + 8t^2 - 6t^4$ .

Step 2: Find the velocity by taking the derivative of the displacement with respect to time: $v = \frac{{dy}}{{dt}} = 3 + 16t - 24t^3$ .

Step 3: To find the initial velocity, set $t = 0$ : $v(0) = 3 + 16(0) - 24(0)^3 = 3$ m/s.

Step 4: Final Answer: $v(0) = 3$ m/s.

Why other options are wrong:
- Option A: This is incorrect as it is a wrong calculation of the initial velocity.
- Option B: This is incorrect as it is a wrong calculation of the initial velocity.
- Option D: This is incorrect as it is a wrong calculation of the initial velocity.

Key Concepts:
- Velocity as the derivative of displacement
- Initial conditions

Memory Aids:
- Velocity is the rate of change of displacement.

Quick Tips:
- Always take the derivative to find velocity from displacement.

26 Easy

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Understanding +4 | -1

A body is thrown vertically upwards with a velocity u. Find the true statement from the following

View answer and explanation

Correct option: D

Velocity is zero at the highest point and maximum height reached is

{u^2}/2g

Option D is correct because at the highest point, the velocity is momentarily zero, and the maximum height is given by

{u^2}/2g

based on kinematic equations.

Solution approach

Correct Answer + Core Reason: Velocity is zero at the highest point and maximum height reached is ${u^2}/2g$ . At the highest point, the body momentarily stops before changing direction, and the maximum height is derived from kinematic equations.

Step-by-Step Solution:

Step 1: Identify the initial conditions: Initial velocity = u, final velocity at the highest point = 0, acceleration = -g (due to gravity).

Step 2: Use the kinematic equation: $v^2 = u^2 + 2as$ , where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement (height).

Step 3: Substitute the values: $0 = u^2 + 2(-g)H$ , where H is the maximum height.

Step 4: Solve for H: $2gH = u^2$ , so $H = \frac{{u^2}}{{2g}}$

Step 5: At the highest point, the velocity is zero as the body momentarily stops before changing direction.

Why other options are wrong:
- Option A: Both velocity and acceleration are zero at its highest point. This is incorrect because while the velocity is zero, the acceleration is always g downwards due to gravity.
- Option B: Velocity is maximum and acceleration is zero at the highest point. This is incorrect because the velocity is minimum (zero) at the highest point, and the acceleration is g downwards.
- Option C: Velocity is maximum and acceleration is g downwards at its highest point. This is incorrect because the velocity is minimum (zero) at the highest point.

Key Concepts:
- Kinematic equations of motion
- Acceleration due to gravity
- Vertical projectile motion

Memory Aids:
- Remember the kinematic equation: $v^2 = u^2 + 2as$
- At the highest point in projectile motion, v=0.

Quick Tips:
- Always consider the direction of acceleration due to gravity (negative if upwards is positive).
- Remember that velocity is a vector, so direction matters.

27 Easy

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Application +4 | -1

The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius

r

is:

View answer and explanation

Correct option: B

\pi / 2

Option B (

\pi / 2

) is correct because the distance is half the circumference of the circle and the displacement is twice the radius.

Solution approach

Correct Answer + Core Reason: $\pi / 2$ . The distance is half the circumference, and the displacement is the diameter.

Step-by-Step Solution:

Step 1: Define the distance covered as half the circumference of the circle: $d = \pi r$ .

Step 2: Define the displacement as the diameter of the circle: $s = 2r$ .

Step 3: Calculate the ratio of distance to displacement: $\frac{d}{s} = \frac{{\pi r}}{{2r}} = \frac{\pi}{2}$ .

Step 4: Final Answer: $\pi / 2$

Why other options are wrong:
- **Option A ( $\pi / 4$ ): This is incorrect because it uses an incorrect formula or makes a mistake in the calculation.
- Option C ( $3 \pi / 4$ ): This is incorrect because it uses an incorrect formula or makes a mistake in the calculation.
- Option D ( $\pi$ ): This is incorrect because it uses an incorrect formula or makes a mistake in the calculation.

Key Concepts:
- Distance and Displacement
- Circumference of a Circle

Memory Aids:
- Distance is the total path length, displacement is the shortest distance between initial and final points.

Quick Tips:**
- Remember the difference between distance and displacement.

28 Easy

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Understanding +4 | -1

The slope of velocity-time graph for motion with uniform velocity is equal to

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because the slope of a velocity-time graph represents acceleration, and for uniform velocity, the acceleration is zero.

Solution approach

Correct Answer + Core Reason: Zero. The slope of a velocity-time graph for uniform velocity is zero.

Step-by-Step Solution:

Step 1: Recall that the slope of a velocity-time graph represents acceleration.

Step 2: For motion with uniform velocity, the velocity is constant, meaning there is no change in velocity over time.

Step 3: Acceleration is defined as the rate of change of velocity: $a = \frac{\Delta v}{\Delta t}$ .

Step 4: Since the velocity is constant, $\Delta v = 0$ , so $a = 0$ .

Step 5: Therefore, the slope of the velocity-time graph is zero.

Why other options are wrong:
- Option A: The slope is not equal to the final velocity; it's equal to the acceleration, which is zero.
- Option B: The slope is not equal to the initial velocity; it's equal to the acceleration, which is zero.
- Option D: Option C is correct.

Key Concepts:
- Velocity-Time Graph
- Slope of a Graph
- Acceleration
- Uniform Velocity

Memory Aids:
- Remember that the slope of a velocity-time graph represents acceleration.

Quick Tips:
- For uniform velocity, the acceleration is always zero.

29 Easy

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Understanding +4 | -1

The velocity of a body at a given instant of time during motion is:

View answer and explanation

Correct option: C

Instantaneous velocity

=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{s}}{\Delta t}

Option C is correct. Instantaneous velocity is defined as the limit of the average velocity as the time interval approaches zero.

Solution approach

Correct Answer + Core Reason: Instantaneous velocity $=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{{s}}}{\Delta t}$ . This is the definition of instantaneous velocity.

Step-by-Step Solution:

Step 1: Recall the definition of average velocity: $v_{avg} = \frac{\Delta s}{\Delta t}$ .

Step 2: Instantaneous velocity is the velocity at a specific instant in time. To find it, we take the limit of the average velocity as the time interval approaches zero: $v = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}$ .

Why other options are wrong:
- Option A: Instantaneous velocity $=\frac{\Delta \vec{{s}}}{\Delta t}$ . This represents average velocity, not instantaneous velocity.
- Option B: Average velocity $=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{{s}}}{\Delta t}$ . This is the definition of instantaneous velocity, not average velocity.
- Option D: None of these. Option C is the correct definition of instantaneous velocity.

Key Concepts:
- Average Velocity
- Instantaneous Velocity
- Limits

Memory Aids:
- Instantaneous = at an instant, so use a limit.

Quick Tips:
- Remember the difference between average and instantaneous quantities. Instantaneous quantities involve limits.

30 Easy

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Understanding +4 | -1

A particle covers equal distance around a circular path, in equal intervals of time. Which of the following quantities connected with the motion of the particle remains constant with time?

View answer and explanation

Correct option: C

Speed

Option C is correct because covering equal distances in equal time intervals implies constant speed.

Solution approach

Correct Answer + Core Reason: Speed. Equal distance covered in equal time intervals means constant speed.

Step-by-Step Solution:
Step 1: Define speed: Speed is the rate at which an object covers distance.
Step 2: Analyze the given condition: The particle covers equal distances in equal intervals of time.
Step 3: Conclude: This implies that the speed of the particle is constant.

Why other options are wrong:
- Option A: Displacement changes as the particle moves along the circular path.
- Option B: Velocity changes because the direction of motion changes, even if the speed is constant.
- Option D: Acceleration is non-zero because the direction of the velocity is changing (centripetal acceleration).

Key Concepts:
- Speed
- Velocity
- Acceleration
- Circular Motion

Memory Aids:
- Speed is a scalar (magnitude only), velocity is a vector (magnitude and direction).
- In uniform circular motion, speed is constant, but velocity is not.

Quick Tips:
- Differentiate between speed and velocity.
- Remember that acceleration can be caused by a change in speed or direction.

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