Skip to main content
Back to MOTION IN A STRAIGHT LINE

Your Progress

0 / 200 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • MOTION IN A STRAIGHT LINE

Motion in a straight line practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 151-180 of 200 questions

Preparing practice and saved progress…

151 Medium
151 of 200
Analyzing +4 | -1
Consider the acceleration, velocity and displacement of a tennis ball as it falls to the ground and bounces back. Directions of which of these changes in the process
View answer and explanation
Correct option: B
Displacement and Velocity
Option B (Displacement and Velocity) is correct because the acceleration due to gravity remains constant in the downward direction, but the velocity and displacement change direction upon bouncing.
152 Easy
152 of 200
Application +4 | -1
If Position of a particle is given by x=(4t2−8t)x=\left(4 t^2-8 t\right), then which of the following is true?
View answer and explanation
Correct option: C
Velocity is zero at t=1 s\mathrm{t}=1 \mathrm{~s}
The correct answer is C (Velocity is zero at t=1 s) because the derivative of the position function, representing velocity, equals zero at t=1.
153 Medium
153 of 200
Understanding +4 | -1
The numerical ratio of displacement to the distance covered is always
View answer and explanation
Correct option: C
Equal to or less than one
Option D (Both statements are false) is correct. Speed is always non-negative, and direction is accounted for by velocity, not speed.
154 Easy
154 of 200
Analyzing +4 | -1
Assertion (A): Speed of a particle may decrease, even if acceleration is increasing.

Reason (R): This will happen if acceleration is positive.
View answer and explanation
Correct option: C
If Assertion is true, but the Reason is false.
Option C (If Assertion is true, but the Reason is false) is correct. The speed of a particle can decrease even if the acceleration is increasing, but this occurs when the acceleration is in the opposite direction to the velocity, not necessarily when acceleration is positive.
155 Medium
155 of 200
Understanding +4 | -1
A body is moving along a straight line path with constant velocity. At an instant of time the distance travelled by it is s and its displacement is D, then
View answer and explanation
Correct option: C
D = S
Option C (D = S) is correct because when a body moves along a straight line path with constant velocity, the distance traveled is equal to the magnitude of the displacement.
156 Medium
156 of 200
Understanding +4 | -1
Statement-I: The speed of a body can be negative.
Statement-II: If the body is moving in the opposite direction of positive motion, then its speed is negative.
View answer and explanation
Correct option: D
Both statements are false
Option C (Equal to or less than one) is correct because the magnitude of displacement is always less than or equal to the distance covered.
157 Easy
157 of 200
Application +4 | -1
A particle moving with uniform acceleration has its final velocity (v=150+8x)  m/s(v = \sqrt {150 + 8x} )\,\,m\mathrm{/}s where xx is the distance travelled by the body then the acceleration
View answer and explanation
Correct option: B
4  m/s24\,\,m\mathrm{/}{s^2}
Option B (4  m/s24\,\,m\mathrm{/}{s^2}) is correct because it is derived by differentiating the given velocity equation with respect to x and using the chain rule.
158 Easy
158 of 200
Application +4 | -1
A car travels equal distances in the same direction with velocities 60kmh−1 , 20  km  h−160\mathrm{km}{\mathrm{h}^{ - 1}}\,,\,20\;\mathrm{km\;}{\mathrm{h}^{ - 1}} and 10  km  h−110\;\mathrm{km\;}{\mathrm{h}^{ - 1}} respectively. The average velocity of the car over the whole journey of motion is
View answer and explanation
Correct option: D
5 ms−15\,\mathrm{m}{\mathrm{s}^{ - 1}}
Option D (5 ms−15\,\mathrm{m}{\mathrm{s}^{ - 1}}) is correct because it represents the harmonic mean of the velocities, converted to m/s.
159 Hard
159 of 200
Application +4 | -1
Two particles are moving with velocities 22 m/sec and
3 m/sec along X-axis in opposite direction. Their initial
positions are at x = 30 m, x = 180 m respectively. find
time interval after which they meet with each other
View answer and explanation
Correct option: C
6 sec
Option C (6 sec) is correct because it is the time calculated by considering the relative velocity and the initial separation between the two particles.
160 Easy
160 of 200
Application +4 | -1
The velocity of light in vaccum is 3×108 m/s3 \times {10^8}\,m/s. Calculate the time taken by it to travel a distance of 6×1014 m6 \times {10^{14}}\,m
View answer and explanation
Correct option: C
t=2×106 st = 2 \times {10^6}\,s
Option C (t=2×106 st = 2 \times {10^6}\,s) is correct because it is calculated by dividing the distance by the speed of light.
161 Medium
161 of 200
Understanding +4 | -1
The numerical ratio of displacement to distance for a moving object is
View answer and explanation
Correct option: D
Equal to or less than 1
Option D (Equal to or less than 1) is correct because the magnitude of displacement is always less than or equal to the distance traveled. The ratio can be 1 only if the object moves in a straight line without changing direction.
162 Medium
162 of 200
Analyzing +4 | -1
A body is projected up with a speed ‘u' and the time taken by it is T to reach the maximum height H. Pick out the correct statement
View answer and explanation
Correct option: B
It acquires velocity u/2 in T/2sec
Option B (It acquires velocity u/2 in T/2sec) is correct because, under constant gravitational acceleration, velocity changes linearly with time. Therefore, half the initial velocity is reached in half the total time to reach maximum height.
163 Medium
163 of 200
Understanding +4 | -1
Stopping distance of a moving vehicle is directly proportional to
View answer and explanation
Correct option: A
Square of the initial velocity
Option A (Square of the initial velocity) is correct because stopping distance is proportional to the square of the initial velocity, as derived from kinematic equations.
164 Medium
164 of 200
Application +4 | -1
A man goes 10 m towards North, then 20 m towards east then displacement is
View answer and explanation
Correct option: A
22.5 m
Option A (22.5 m) is incorrect. The correct answer should be 22.36 m, which is not listed. The closest option is A, but it is rounded.
165 Medium
165 of 200
Analyzing +4 | -1
A body starts from rest and travels a distance S with uniform acceleration, then moves uniformly a distance 2 S uniformly, and finally comes to gest after moving further 5 S under uniform retardation. The ratio of average velocity to maximum velocity is
View answer and explanation
Correct option: C
47\frac{4}{7}
Option C (47\frac{4}{7}) is correct because it is derived by calculating the average and maximum velocities and then finding their ratio.
166 Easy
166 of 200
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 final velocity is equal in magnitude but opposite in direction to the initial velocity when a body returns to its starting point under gravity.
167 Hard
167 of 200
Application +4 | -1
Amongst the following equation of motion, which represents uniformly accelerated motion?
View answer and explanation
Correct option: C
t=x+abt=\sqrt{\frac{x+a}{b}}
Option C (t=x+abt=\sqrt{\frac{{x+a}}{{b}}}) is correct because solving for x yields an equation of the form x=bt2−ax = bt^2 - a, which represents uniformly accelerated motion.
168 Medium
168 of 200
Analyzing +4 | -1
In one dimensional motion, instantaneous speed vv satisfies 0≤v<v00 \leq v < v_0.
View answer and explanation
Correct option: B
The displacement xx in time TT satisfies
−v0T<x<v0T-v_0 T < x < v_0 T
Option B is correct because it provides the correct range for displacement based on the maximum speed and time interval.
169 Medium
169 of 200
Remembering +4 | -1
Which branch of physics deals with motion without considering its causes?
View answer and explanation
Correct option: C
Kinematics
Option C (Kinematics) is correct because kinematics specifically deals with the description of motion without considering the forces that cause it.
170 Medium
170 of 200
Application +4 | -1
A train started from rest from a station and accelerated at 2 ms−22\,m{s^{ - 2}} for 10sec. Then it travels at constant speed for 30 sec and there after it decelerated at 4 ms−24\,m{s^{ - 2}} until it stopped at the next station. The distance between two stations is
View answer and explanation
Correct option: C
750 m
Option C (750 m) is correct because it is the total distance calculated by summing the distances traveled during acceleration, constant speed, and deceleration.
171 Medium
171 of 200
Analyzing +4 | -1
The relative velocity VAB\mathrm{V}_{\mathrm{AB}} or VBA\mathrm{V}_{\mathrm{BA}} of two bodies A\mathrm{A} and B\mathrm{B} may be

I. greater than velocity of body A

II. greater than velocity of body B

III. less than the velocity of body A

IV. less than the velocity of body B
View answer and explanation
Correct option: D
I, II, III and IV
I, II, III and IV. The relative velocity can be greater or less than the individual velocities depending on the direction of motion.
172 Easy
172 of 200
Application +4 | -1
The x and y coordinates of a particle at any time tt are given by x=8t2+5t3x = 8{t^2} + 5{t^3} and y=2ty = 2t. Where x and y are in metre and tt in seconds. The acceleration of particle at t=5t = 5 sec is
View answer and explanation
Correct option: D
166 m/s2166\,m/{s^2}
Option D (166 m/s2166\,m/{s^2}) is correct because it is the magnitude of the acceleration vector calculated by finding the second derivatives of the x and y coordinates with respect to time and evaluating at t=5s.
173 Hard
173 of 200
Analyzing +4 | -1
The distance x covered by a particle in one-dimensional motion varies with time t as x2=at2+2bt+c{x^2} = a{t^2} + 2bt + c. The accelerating of the particle varies as
View answer and explanation
Correct option: A
x−3{x^{ - 3}}
Option A (x−3{x^{ - 3}}) is correct because the acceleration is inversely proportional to the cube of the distance x, derived from differentiating the given equation.
174 Easy
174 of 200
Application +4 | -1
A police jeep is chasing with velocity of 45 km/h45 \mathrm{~km} / \mathrm{h} a thief in another jeep moving with velocity 153 km/h153 \mathrm{~km} / \mathrm{h}. Police fires a bullet with muzzle velocity of 180 m/ s180 \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 m/s150 \mathrm{~m} / \mathrm{s}
Option A (150 m/s) is correct because it calculates the relative velocity of the bullet with respect to the thief's car.
175 Easy
175 of 200
Analyzing +4 | -1
The velocity v\mathrm{v} of a particle is given by the equation v=6t2−6t3v=6 t^2-6 t^3 where vv is in m/sec\mathrm{m} / \mathrm{sec} and t\mathrm{t} is time in seconds then :
View answer and explanation
Correct option: C
minimum velocity is zero
Option D (minimum velocity is zero) is correct because the minimum velocity occurs when the derivative of the velocity function is zero, leading to a zero velocity value.
176 Medium
176 of 200
Understanding +4 | -1
A graph is shown between velocity an time for the motion of a particle. The area under the curve between the time intervals t1 and t2{t_1}\,and\,{t_2} gives
View answer and explanation
Correct option: B
Displacement of the particle
Option B (Displacement of the particle) is correct because the area under the velocity-time graph represents the displacement of the particle.
177 Hard
177 of 200
Application +4 | -1
Two bodies move in a straight line towards each other with initial velocities v1v_1 and v2v_2 and with constant accelerations a1a_1 and a2a_2 directed against their corresponding initial velocities. The maximum initial separation ℓmax⁡\ell_{\max } between them for which they will meet during the motion is:
View answer and explanation
Correct option: B
(v1+v2)22(a1+a2)\frac{\left(v_1+v_2\right)^2}{2\left(a_1+a_2\right)}
Option B ((v1+v2)22(a1+a2)\frac{\left(v_1+v_2\right)^2}{2\left(a_1+a_2\right)}) is correct because it represents the maximum initial separation for the two bodies to meet, considering their initial velocities and decelerations.
178 Medium
178 of 200
Understanding +4 | -1
Assertion : Displacement of a body is vector sum of the area under velocity- time graph.

Reason : Displacement is a vector quantity.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the area under the velocity-time graph represents displacement, and displacement is a vector quantity, making the vector sum relevant.
179 Hard
179 of 200
Application +4 | -1
A ball is thrown vertically upwards. It was observed at a height h twice with a time interval Δ\Deltat. The initial velocity of the ball is
View answer and explanation
Correct option: C
128gh+g2(Δt)2\frac{1}{2}\sqrt {8\mathrm{g}h + {\mathrm{g}^2}{{\left( {\Delta t} \right)}^2}}
Option C (128gh+g2(Δt)2\frac{1}{2}\sqrt {8\mathrm{g}h + {\mathrm{g}^2}{{\left( {\Delta t} \right)}^2}}) is correct because it is derived from the equations of motion, considering the time interval between the ball passing the height h twice.
180 Medium
180 of 200
Understanding +4 | -1
A graph is drawn between velocity and time for the motion of a particle. The area under the curve between the time intervals t1 and  t2{t_1}\,and\,\,{t_2} gives
View answer and explanation
Correct option: B
Displacement of the particle
Option B (Displacement of the particle) is correct because the area under a velocity-time graph represents the displacement of the particle.
Revision notes and key takeaways

Review MOTION IN A STRAIGHT LINE in your textbook or class notes, then use this set to check your understanding.

Showing 151-180 of 200 questions

Your practice

0 / 200 answered · 200 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.