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JEE • PHYSICS • MOTION IN A STRAIGHT LINE

Motion in a straight line practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 285 questions

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151 Medium
151 of 285
Application +4 | -1
The position of a particle moving along xx-axis varies with time tt as x=4t−t2+1x=4 t-t^2+1. Find the time interval(s) during which the particle is moving along positive xx-direction.
View answer and explanation
Correct option: C
0<t<20 < \mathrm{t} < 2.
Option C (0 < t < 2) is correct because the particle moves in the positive x-direction when its velocity (dx/dt) is positive, which occurs for t < 2.
152 Medium
152 of 285
Analyzing +4 | -1
A point moves such that its displacement as a function of time is given by x3=t3+1x^3=t^3+1. Its acceleration as a function of time tt will be
View answer and explanation
Correct option: B
2tx5\frac{2 t}{x^5}
Option B (2tx5\frac{2 t}{x^5}) is correct because it is the result of differentiating the given equation twice with respect to time to find the acceleration in terms of x and t.
153 Medium
153 of 285
Understanding +4 | -1
Assertion: The average velocity of the object over an interval of time is either smaller than or equal to the average speed of the object over the same interval.

Reason: Velocity is a vector quantity and speed is a scalar 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 average velocity's magnitude is always less than or equal to the average speed due to displacement being less than or equal to distance, and this is because velocity is a vector and speed is a scalar.
154 Medium
154 of 285
Application +4 | -1
A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three second. The body has fallen for a time of:-
View answer and explanation
Correct option: B
5 s5\,s
Option B (5 s) is correct because the time is calculated by equating the distance traveled in the first 3 seconds to the distance traveled in the last second.
155 Medium
155 of 285
Analyzing +4 | -1
A body projected vertically upwards with a certain speed from top of a tower, reaches the ground in time t1.t_1. If it is projected vertically downwards with same speed from same point, it reach the ground in the time t2.t_2. Time required for it to reach the ground if it dropped from the same point and projected horizontally with same speed, respectively are
View answer and explanation
Correct option: D
t1t2,t1t2\sqrt{t_1 t_2}, \sqrt{t_1 t_2}
The correct answer is D (t1t2,t1t2\sqrt{t_1 t_2}, \sqrt{t_1 t_2}) because the time to fall when dropped is the geometric mean of the times when projected upwards and downwards, and the time for horizontal projection is the same as dropping.
156 Medium
156 of 285
Application +4 | -1
The ball is projected up from ground with speed 30 m/sec. What is the average velocity for time 0 to 4 sec?
View answer and explanation
Correct option: A
10 m/sec
Option A (10 m/sec) is correct because it is calculated by finding the displacement of the ball after 4 seconds and dividing it by the time interval.
157 Medium
157 of 285
Understanding +4 | -1
Assertion: The slope of the velocity-time graph for an object in uniform motion is zero.

Reason: The velocity-time graph for uniform motion is parallel to velocity-axis.
View answer and explanation
Correct option: C
If assertion is true, but reason is false.
Option C is correct because the slope of a velocity-time graph represents acceleration, which is zero for uniform motion, but the graph is parallel to the time-axis, not the velocity-axis.
158 Medium
158 of 285
Application +4 | -1
A car moving on straight path, travels first half of distance with speed 20  m/s20\, {~m} / {s} and another half of distance with speed 40  m/s40 \,{~m} / {s}, then average speed of the journey is
View answer and explanation
Correct option: B
803  m/s\frac{80}{3}\, {~m} / {s}
Option B (803 m/s\frac{80}{3}\, {\text{{m}}} / {\text{{s}}}) is correct because it calculates the average speed considering different speeds over equal distances.
159 Medium
159 of 285
Application +4 | -1
A wheel covers a distance of 9.5 km9.5 \mathrm{~km} in 2000 revolutions. The diameter of the wheel is
View answer and explanation
Correct option: C
1.5 m1.5 \mathrm{~m}
Option C (1.5 m1.5 \mathrm{~m}) is correct because it is calculated using the relationship between distance, number of revolutions, and circumference of the wheel.
160 Medium
160 of 285
Understanding +4 | -1
In the above question, the nearest distance between the two persons is
View answer and explanation
Correct option: C
8 m
Option C (8 m) is correct because it represents the minimum distance calculated by minimizing the distance function with respect to time.
161 Medium
161 of 285
Application +4 | -1
For motion of an object along xx-axis, the velocity vv depends on the position xx as v=(2x2−x)m/sv=\left(2 x^2-x\right) \mathrm{m} / \mathrm{s}, where xx is in m\mathrm{m}. The acceleration of the object at x=1 mx=1 \mathrm{~m} will be
View answer and explanation
Correct option: B
3 m/s23 \mathrm{~m} / \mathrm{s}^2
Option B (3 m/s23 \mathrm{~m} / \mathrm{s}^2) is correct because it is the result of applying the chain rule to find acceleration as a=v(dv/dx)a = v(dv/dx) and evaluating at x=1 mx=1 \mathrm{~m}.
162 Medium
162 of 285
Application +4 | -1
A stone thrown vertically upwards with a speed of 5 m/sec5 \mathrm{~m} / \mathrm{sec} attains a height H1H_1. Another stone thrown upwards from the same point with speed of 10 m/sec10 \mathrm{~m} / \mathrm{sec} attains a height H2H_2. The correct relation between H1H_1 and H2H_2 is:
View answer and explanation
Correct option: A
H2=4H1{H}_2=4 H_1
Option A (H2=4H1{H}_2=4 H_1) is correct because the maximum height is proportional to the square of the initial velocity, and the second stone's initial velocity is twice that of the first.
163 Medium
163 of 285
Application +4 | -1
A stone is thrown with an initial speed of 4.9 m/s from a bridge in vertically upward direction. It falls down in water after 2 sec. The height of the bridge is
View answer and explanation
Correct option: B
9.8 m
The correct answer is B (9.8 m). The height of the bridge is calculated using the kinematic equation for vertical motion under gravity, considering the initial upward velocity and the total time of flight.
164 Medium
164 of 285
Analyzing +4 | -1
The displacement x of a particle varies with time according
to the relation ) x=ab(1−e−bt)x = \frac{a}{b}\left( {1 - {e^{ - bt}}} \right). Then select the false alternative.
View answer and explanation
Correct option: D
The particle will not come back to its starting point at t→∞t \to \infty
Option D (The particle will not come back to its starting point at t→∞t \to \infty) is correct because as t approaches infinity, x approaches a/b, which is not the starting point (x=0).
165 Medium
165 of 285
Application +4 | -1
A body moves in straight line with velocity v1v_1 for 1/3rd 1 / 3^{\text {rd }} time and for remaining time with v2\mathrm{v}_2. Find average velocity.
View answer and explanation
Correct option: A
v13+2v23\frac{\mathrm{v}_1}{3}+\frac{2 \mathrm{v}_2}{3}
Option A (v13+2v23\frac{\mathrm{v}_1}{3}+\frac{2 \mathrm{v}_2}{3}) is correct because it is the weighted average of the velocities based on the time intervals.
166 Medium
166 of 285
Application +4 | -1
A particle travels 80 m with 5 m/s velocity, later it
continuous to travel 20 m in 9 sec. Find average
velocity
View answer and explanation
Correct option: D
4 m/s
Option D (4 m/s) is correct because the average velocity is calculated by dividing the total displacement by the total time.
167 Medium
167 of 285
Application +4 | -1
A particle moves in straight line with velocity v1v_1 and v2v_2 for time intervals which are in ratio 1:2.1:2. Find average velocity.
View answer and explanation
Correct option: A
v13+2v23\frac{\mathrm{v}_1}{3}+\frac{2 \mathrm{v}_2}{3}
Option A (v13+2v23\frac{\mathrm{v}_1}{3}+\frac{2 \mathrm{v}_2}{3}) is correct because it is the average velocity calculated considering the time intervals.
168 Medium
168 of 285
Application +4 | -1
A Body moves 6 m north. 8 m east and 10 m vertically upwards, what is its resultant displacement from initial position
View answer and explanation
Correct option: A
102m10\sqrt {2m}
Option A (102m10\sqrt {2m}) is correct because it calculates the magnitude of the resultant displacement vector using the Pythagorean theorem in 3D.
169 Medium
169 of 285
Application +4 | -1
A particle starts from rest with constant acceleration. The ratio of magnitude of displacement in nth n^{\text {th }} second to that of in nn second, is
View answer and explanation
Correct option: D
(2n−1n2)\left(\frac{2}{n}-\frac{1}{n^2}\right)
Option D is correct because the ratio of displacement in the nth second to that in n seconds is (2/n - 1/n²), derived from kinematic equations.
170 Medium
170 of 285
Application +4 | -1
The relation between tt and distance xx is t=ax2+bxt=a x^2+b x where aa and bb are constants. Express the instantaneous acceleration in terms of instantaneous velocity.
View answer and explanation
Correct option: A
−2av3-2 a v^3
Option A (−2av3-2 a v^3) is correct because it is the result of differentiating the given relation with respect to time and expressing acceleration in terms of velocity.
171 Medium
171 of 285
Analyzing +4 | -1
A body is thrown up with a velocity 100 ms−1100 \mathrm{~ms}^{-1}. It travels 5 m5 \mathrm{~m} in the last second of its journey. If the same body is throw up with a velocity 200 ms−1200 \mathrm{~ms}^{-1}, how much distance (in metre will it travel in the last second (g=10 ms−2)\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right) ?
View answer and explanation
Correct option: A
5 m
Option A (5 m) is correct because the distance traveled in the last second of upward motion is independent of the initial velocity and depends only on gravity.
172 Medium
172 of 285
Analyzing +4 | -1
Two bodies of different masses are dropped simultaneously from the top of a tower. If air resistance is proportional to the mass of the body. Then
View answer and explanation
Correct option: C
Both the bodies reach the ground simultaneously
Option C is correct because if air resistance is proportional to the mass, the effective acceleration is the same for both bodies, leading to simultaneous arrival.
173 Medium
173 of 285
Analyzing +4 | -1
Assertion : position-time graph of a body moving uniformly in a straight line parallel to position axis. Says body is at rest.

Reason : The slope of position-time graph in a uniform motion gives the velocity of an object.
View answer and explanation
Correct option: C
Assertion is correct, reason is incorrect
Option D (Assertion is incorrect, reason is correct) is correct because a position-time graph parallel to the position axis indicates infinite velocity, and the slope of a position-time graph gives the velocity.
174 Medium
174 of 285
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
Option D (I, II, III and IV) is correct because the relative velocity can be greater or less than the individual velocities depending on the direction of motion.
175 Medium
175 of 285
Application +4 | -1
From the top of tower, a particle is thrown vertically downwards with a velocity of 10 m/s10 \mathrm{~m} / \mathrm{s}. The ratio of the distances covered by it in the 3rd and 2nd seconds of the motion is : (Take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 ) :
View answer and explanation
Correct option: B
0.295138889
Option B (07:05) is correct because it is the ratio of distances calculated using the formula for distance covered in the nth second.
176 Medium
176 of 285
Understanding +4 | -1
The area under acceleration-time graph gives
View answer and explanation
Correct option: C
Change in velocity
Option C (Change in velocity) is correct because the area under the acceleration-time graph represents the integral of acceleration with respect to time, which is the change in velocity.
177 Medium
177 of 285
Understanding +4 | -1
Assertion: The average speed of a body over a given interval of time is equal to the average velocity of the body in the same interval of time if a body moves in a straight line in one direction.
Reason: Because in this case distance travelled by a body is equal to the displacement of the body.
View answer and explanation
Correct option: A
If both assertion and reason are true and the reason is the correct explanation of the assertion.
Option A is correct because when a body moves in a straight line in one direction, the distance traveled equals the displacement, making average speed and average velocity equal.
178 Medium
178 of 285
Application +4 | -1
A stone is dropped from a rising balloon at a height of 76 m76 \mathrm{~m} above the ground and reaches the ground in 6 s6 \mathrm{~s}. What was the velocity of the balloon when the stone was dropped? [Take g=10 m/s2{g}=10 \mathrm{~m} / \mathrm{s}^2]
View answer and explanation
Correct option: A
(523)m/s\left(\frac{52}{3}\right) \mathrm{m} / \mathrm{s} upwards
Option A (52/3 m/s upwards) is correct because it is the initial upward velocity of the stone calculated using the kinematic equation with the given height, time, and acceleration due to gravity.
179 Medium
179 of 285
Understanding +4 | -1
The location of a particle has changed. What can we say about the displacement and the distance covered by the particle?
View answer and explanation
Correct option: A
Neither can be zero
Option A (Neither can be zero) is correct because if the location of a particle has changed, it must have covered some distance and undergone some displacement.
180 Medium
180 of 285
Application +4 | -1
The displacement of the particle varies with time according to the relation x=kb[1−e−bt].x = \frac{k}{b}\left[ {1 - {e^{ - bt}}} \right]. Then the velocity of the particle is
View answer and explanation
Correct option: A
k(e−bt)k\left( {{e^{ - bt}}} \right)
Option A (k(e−bt)k(e^{-bt})) is correct because it is the derivative of the given displacement function with respect to time, representing the particle's velocity.
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