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JEE • PHYSICS • MOTION IN A STRAIGHT LINE

Motion in a straight line practice · Mixed difficulty

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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 121-150 of 333 questions

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121 Medium
121 of 333
Application +4 | -1
Two trains, each of length 100 m100 \mathrm{~m}, are running on parallel tracks. One overtakes the other in 20sec20 \mathrm{sec}. and one crosses the other in opposite direction 10 sec. Calculate the velocities of each train.
View answer and explanation
Correct option: A
15 m/s15 \mathrm{~m} / \mathrm{s} and 5 m/s5 \mathrm{~m} / \mathrm{s}
Option A (15 m/s15 \text{{ m/s}} and 5 m/s5 \text{{ m/s}}) is correct because it solves the system of equations derived from relative velocities in overtaking and crossing scenarios.
122 Medium
122 of 333
Application +4 | -1
A point moves in a straight line, so that its displacement xx at time tt is given by x2=t2+1x^2=t^2+1. Its acceleration is
View answer and explanation
Correct option: B
1/x31 / x^3
Option B (1/x31 / x^3) is correct because it is derived from differentiating the displacement equation twice with respect to time.
123 Hard
123 of 333
Application +4 | -1
A particle moves according to the equation dvdt=α−βv\frac{d v}{d t}=\alpha-\beta v, where α\alpha and β\beta are constants. Find velocity as a function of time. Assume, body starts from rest
View answer and explanation
Correct option: B
v=(α/β)(1−e−βt)v=(\alpha / \beta)\left(1-e^{-\beta t}\right)
Option B (v=(α/β)(1−e−βt)v=(\alpha / \beta)(1-e^{-\beta t})) is correct because it is the solution to the given differential equation with the initial condition that the particle starts from rest.
124 Medium
124 of 333
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 displacement can be equal to or less than the distance traveled.
125 Medium
125 of 333
Understanding +4 | -1
Two bodies are held separated by 9.8 m9.8 \mathrm{~m} vertically ane above the other. They are released simultaneously to fall freely under gravity. After 2 s2 \mathrm{~s}, the relative distance between them is
View answer and explanation
Correct option: C
9.8 m9.8 \mathrm{~m}
Option C (9.8 m9.8 \mathrm{~m}) is correct because both bodies experience the same acceleration due to gravity, maintaining their initial relative distance.
126 Medium
126 of 333
Understanding +4 | -1
The numerical ratio of displacement to distance is
View answer and explanation
Correct option: D
may be less than 1 or equal to one
Option D (may be less than 1 or equal to one) is correct because displacement can be less than or equal to the distance traveled, but never greater.
127 Medium
127 of 333
Understanding +4 | -1
Velocity – time graph of a body thrown vertically up is
View answer and explanation
Correct option: A
a straight line
Option A (a straight line) is correct because the velocity of a body thrown vertically upwards changes linearly with time due to constant gravitational acceleration.
128 Medium
128 of 333
Application +4 | -1
If V=αx,VV=\alpha \sqrt{x}, \mathbf{V} is velocity, x\mathbf{x} is position, a\mathbf{a} is +ve constant then a=\mathbf{a}= ?
View answer and explanation
Correct option: D
a=α22a=\frac{\alpha^2}{2}
Option D (a=α22a=\frac{\alpha^2}{2}) is correct because the acceleration is derived by differentiating the velocity equation with respect to position and time.
129 Easy
129 of 333
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 −2 m/sec-2 \mathrm{~m} / \mathrm{sec}) is incorrect. The minimum velocity is zero, occurring at t=0.
130 Medium
130 of 333
Application +4 | -1
A particle moves in a straight line as s=α(t−4)+β(t−4)2s = \alpha \left( {t - 4} \right) + \beta {\left( {t - 4} \right)^2}. Find the initial velocity and acceleration.
View answer and explanation
Correct option: B
2β2\beta
Option B (2β2\beta) is correct because the initial velocity is α\alpha and the acceleration is 2β2\beta, which is constant.
131 Medium
131 of 333
Application +4 | -1
A particle is moving in a straight line. The variation of position 'xx' as a function of time 'tt' is given as x=(t3−6t2+20t+15)mx=\left(t^3-6 t^2+20 t+15\right) {m}. The velocity of the body when its acceleration becomes zero is
View answer and explanation
Correct option: C
8 m/s8 {~m} / {s}
Option C (8 m/s8 {~m} / {s}) is correct because it is the velocity calculated when the acceleration, derived from the position function, is zero.
132 Medium
132 of 333
Understanding +4 | -1
Which of the following statements, is incorrect regarding motion of particle along straight line?
View answer and explanation
Correct option: D
v−tv-t graph is parabola if motion is uniformly accelerated
Option D (v−tv-t graph is parabola if motion is uniformly accelerated) is correct because for uniformly accelerated motion, the v−tv-t graph is a straight line, not a parabola.
133 Easy
133 of 333
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.
134 Medium
134 of 333
Analyzing +4 | -1
Statement-I: In successive time intervals if the average velocities of a particle are equal then the particle must be moving with constant velocity.
Statement-II: When a particle moves with uniform velocity, its displacement always increases with time.
View answer and explanation
Correct option: D
Both statements are false
Option D (Both statements are false) is correct because average velocities can be equal without constant velocity, and displacement can be zero with uniform velocity.
135 Easy
135 of 333
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.
136 Medium
136 of 333
Understanding +4 | -1
What determines the nature of the path followed by the particle
View answer and explanation
Correct option: D
Both (2) and (3 )
Option D (Both (2) and (3 )) is correct because both velocity and acceleration determine the path followed by a particle.
137 Hard
137 of 333
Application +4 | -1
A body PP is thrown vertically up with velocity 30 m/s and another body QQ is thrown up along the same vertically line with the same velocity but 1 second later from the ground. When they meet (g=10 m/s2)\left( {g = 10\,m/{s^2}} \right)
View answer and explanation
Correct option: C
P travels for 3.5 s
Option C (P travels for 3.5 s) is correct because it's calculated by finding the time when the two projectiles meet, considering the 1-second delay.
138 Medium
138 of 333
Understanding +4 | -1
A man leaves his house for a cycle ride. He comes back to his house after half-an-hour after covering a distance of one km\mathrm{km}. What is his average velocity for the ride?
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because average velocity is defined as total displacement divided by total time, and in this case, the displacement is zero.
139 Medium
139 of 333
Application +4 | -1
A particle is thrown vertically upward with speed 45 m/s. Find the distance covered by the particle in the 5th second of its journey.
View answer and explanation
Correct option: B
2.5 m
Option B (2.5 m) is correct because it calculates the distance traveled during the 5th second, considering the particle's upward and downward motion.
140 Easy
140 of 333
Understanding +4 | -1
If a body is moving along a straight path then which of the following is true?
View answer and explanation
Correct option: C
Distance may be greater than the magnitude of displacement
Option A (Distance is always equal to magnitude of displacement) is correct because when a body moves along a straight path without changing direction, the distance traveled is equal to the magnitude of the displacement.
141 Medium
141 of 333
Application +4 | -1
If f=f0[1−tT]f=f_0\left[1-\frac{t}{T}\right], where f=f= acceleration at time tt then find value of vv at t=tt=t. when t=0,u=0\mathbf{t}=\mathbf{0}, \mathbf{u}=\mathbf{0}
View answer and explanation
Correct option: C
v=f0T2v=\frac{f_0 T}{2}
Option C (v=f0T2v=\frac{f_0 T}{2}) is correct because it is the velocity calculated by integrating the acceleration function with respect to time and evaluating at t=T.
142 Medium
142 of 333
Application +4 | -1
A particle starts from rest at the point A(1,2,−3)mA(1,2,-3) {m} and reaches at the point B(3,4,5)mB(3,4,5) {m}. The magnitude of displacement of man is
View answer and explanation
Correct option: D
8.5 m8.5 {~m}
Option D (8.5 m8.5 {~m}) is correct because it correctly calculates the magnitude of the displacement vector between the initial and final points.
143 Medium
143 of 333
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 deals with the motion of objects without considering the forces that cause the motion.
144 Medium
144 of 333
Application +4 | -1
A ball is thrown vertically upwards with a speed of 20 m/s from the top of a tower 250 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them mm reaching the ground is
View answer and explanation
Correct option: C
4 s
Option C (4 s) is correct because the time difference depends only on the initial upward velocity and the acceleration due to gravity.
145 Medium
145 of 333
Application +4 | -1
A plane travelling north at 200 m/s200 \mathrm{~m} / \mathrm{s} turns and then travels south at 200 m/s200 \mathrm{~m} / \mathrm{s}. Its change in velocity is:
View answer and explanation
Correct option: C
400 m/s400 \mathrm{~m} / \mathrm{s} south
Option C (400 m/s400 \mathrm{~m} / \mathrm{s} south) is correct because change in velocity is final velocity minus initial velocity, considering direction.
146 Medium
146 of 333
Application +4 | -1
A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 254ms−1254\mathrm{m}{\mathrm{s}^{ - 1}}. The two stones will meet after
View answer and explanation
Correct option: A
4s
Option A (4s) is correct because it is the time at which the two stones meet, calculated using kinematic equations.
147 Medium
147 of 333
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 the velocity decreases linearly with time under constant acceleration due to gravity.
148 Medium
148 of 333
Application +4 | -1
A particle under the action of a constant force moves from rest upto 20 seconds. If distance covered in first 10 seconds is S1\mathrm{S}_1 and that covered in next 10 seconds in S2\mathrm{S}_2, then
View answer and explanation
Correct option: B
S2=3 S1\mathrm{S}_2=3 \mathrm{~S}_1
Option B (S2=3 S1\mathrm{S}_2=3 \mathrm{~S}_1) is correct because the distance covered in the next 10 seconds is three times the distance covered in the first 10 seconds.
149 Medium
149 of 333
Analyzing +4 | -1
A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is
View answer and explanation
Correct option: B
Less than the time of fall
Option B is correct because when air resistance is considered, the upward acceleration is greater than the downward acceleration, resulting in a shorter time of ascent compared to the time of descent.
150 Medium
150 of 333
Analyzing +4 | -1
Three ships A, B and C are in motion. The motion of A as seen by B is with speed V towards the north-east. The motion of B as seen by C is with speed V towards the north-west. Then, as seen by A, C will be moving towards the
View answer and explanation
Correct option: B
South
The correct answer is B (South) because by vector addition of relative velocities, the velocity of C as seen by A is towards the South.
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