Skip to main content
Back to MOTION IN A STRAIGHT LINE

Your Progress

0 / 333 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • 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

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.

Take a chapter quiz →
Showing 61-90 of 333 questions

Preparing practice and saved progress…

61 Medium
61 of 333
Application +4 | -1
From the top of a tower, a stone is thrown up and reaches the ground in time t1=9s.{t_1} = 9\mathrm{s}.. A second stone is thrown down with the same speed and reaches the ground in time t2=4{t_2} = 4 s. A third stone is released from rest and reaches the ground in time t3{t_3}, which is equal to
View answer and explanation
Correct option: B
6.0 s
Option C (μ0iD2πr\frac{{{\mathrm{\mu }_0}{i_D}}}{{2\mathrm{\pi }r}}) is correct because it gives the magnetic field outside the capacitor plates (r > R) based on Ampere's law with the displacement current.
62 Medium
62 of 333
Analyzing +4 | -1
The velocity of a particle depends upon the time t\mathrm{t} according to the equation: v=ab+bt+cd+t\mathrm{v}=\sqrt{\mathrm{a b}}+\mathrm{b t}+\frac{\mathrm{c}}{\mathrm{d+t}}. The physical quantities which are represented by a,b,c\mathrm{a, b, c} and d\mathrm{d}, are in the following order:
View answer and explanation
Correct option: B
Distance, acceleration, distance, time
Option B (Distance, acceleration, distance, time) is correct because dimensional analysis of the given equation reveals the physical quantities represented by a, b, c, and d.
63 Medium
63 of 333
Understanding +4 | -1
Can an object maintain uniform velocity when its acceleration is not zero?
View answer and explanation
Correct option: A
No, impossible
Option A is correct because uniform velocity implies zero acceleration; thus, an object cannot maintain uniform velocity if its acceleration is non-zero.
64 Medium
64 of 333
Application +4 | -1
A body travels a distance s1s_1 with velocity v1v_1 and distance s2s_2 with velocity v2v_2 in the same direction. Calculate the average velocity of the body:
View answer and explanation
Correct option: A
(s1+s2)v1v2s1v2+s2v1\frac{\left(s_1+s_2\right) v_1 v_2}{s_1 v_2+s_2 v_1}
Option A ((s1+s2)v1v2s1v2+s2v1\frac{{\left( s_1+s_2 \right) v_1 v_2}}{{s_1 v_2+s_2 v_1}}) is correct because it represents the total distance divided by the total time taken, which is the definition of average velocity.
65 Medium
65 of 333
Application +4 | -1
A ball kicked vertically up attains a height of 19.6m and returns to the point of throw. If the ball is in air for four second, then the value of acceleration due to gravity is
View answer and explanation
Correct option: B
9.8 ms−2\mathrm{9}\mathrm{.8}\,\mathrm{m}{\mathrm{s}^{ - 2}}
Option B (9.8 ms−2\mathrm{9}\mathrm{.8}\,\mathrm{m}{\mathrm{s}^{ - 2}}) is correct because the acceleration due to gravity can be calculated from the time of flight and height reached.
66 Medium
66 of 333
Application +4 | -1
The position vector of particle changes with time according to the relation r(t)=15t2i^+(4−20t2)j^r(t) = 15{t^2}\hat i + (4 - 20{t^2})\hat j. What is the magnitude of the acceleration (in ms−2m{s^{ - \mathrm{2}}}) at t =1 ?
View answer and explanation
Correct option: A
50
Option A (50) is correct because the magnitude of acceleration is calculated by finding the second derivative of the position vector and then taking the magnitude at t=1.
67 Easy
67 of 333
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.
68 Medium
68 of 333
Analyzing +4 | -1
A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 15 S, then
View answer and explanation
Correct option: C
S=172ft2S = \frac{1}{{72}}f{t^2}
Option C (S=172ft2S = \frac{1}{{72}}f{t^2}) is correct because it is derived from kinematic equations and the total distance equation given the acceleration, constant speed, and deceleration phases.
69 Medium
69 of 333
Application +4 | -1
Two trains take 3 sec to pass one another when going in the opposite direction but only 2.5 sec if the speed of the one is increased by 50%50 \%. The time one would take to pass the other when going in the same direction at their original speed is:
View answer and explanation
Correct option: C
15 sec
Option C (15 sec) is correct because by setting up equations for relative speeds in opposite and same directions, we can solve for the individual speeds and then find the time to pass in the same direction.
70 Medium
70 of 333
Analyzing +4 | -1
Assertion: When a body is subjected to a uniform acceleration, it always move in a straight line.

Reason: Straight line motion is the natural tendency of the body.
View answer and explanation
Correct option: D
If assertion is false but reason is true
Option D (If assertion is false but reason is true) is correct because uniform acceleration does not guarantee straight-line motion (e.g., projectile motion), but the tendency of a body to maintain its state is inertia, not necessarily straight-line motion.
71 Medium
71 of 333
Analyzing +4 | -1
STATEMENT-1 : For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.

STATEMENT-2 : If the observer and the object are moving at velocities v⃗1\vec{v}_1 and v⃗2\vec{v}_2 respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is v⃗2−v⃗1\vec{v}_2-\vec{v}_1.
View answer and explanation
Correct option: B
Statement 1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Option B (Statement 1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1) is correct because both statements are true, but the relative velocity formula doesn't directly explain the perceived motion of nearby and distant objects.
72 Medium
72 of 333
Analyzing +4 | -1
Assertion: Kinematic equations are exact and are always correct.
Reason: The definitions of instantaneous velocity and acceleration are true only for motion in which the magnitude and direction of acceleration are constant during motion.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D is correct. Kinematic equations are only correct for constant acceleration. Instantaneous velocity and acceleration are always correct by definition.
73 Medium
73 of 333
Understanding +4 | -1
Assertion : A body can have acceleration even if its velocity is zero at a given instant of time.

Reason : A body is momentarily at rest when it reverses its direction of motion.
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 a body can have zero velocity and non-zero acceleration at an instant, and the reason correctly explains why.
74 Medium
74 of 333
Application +4 | -1
A body travels the first half of the total distance with velocity v1v_1 and the second half with velocity v2.v_2. Calculate the average velocity.
View answer and explanation
Correct option: A
2v1v2v1+v2\frac{2 v_1 v_2}{v_1+v_2}
Option A (2v1v2v1+v2\frac{2 v_1 v_2}{v_1+v_2}) is correct because the average velocity is calculated by dividing the total distance by the total time taken.
75 Medium
75 of 333
Understanding +4 | -1
Assertion: The distance traversed, during equal intervals of time, by a body falling from rest are in ratio 1:3:5:7…1 : 3 : 5 : 7 \ldots [Galileo's law of odd numbers]
Reason: A particle in one-dimensional motion with zero speed may have non-zero velocity.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct. The assertion is true based on Galileo's law of odd numbers, but the reason is false as a particle with zero speed can still have non-zero acceleration.
76 Medium
76 of 333
Analyzing +4 | -1
A particle is projected vertically from the ground takes time t1=1 st_1=1 \mathrm{~s} upto point A,  t2=3 s{A},\,\,t_2=3 \mathrm{~s} from point A{A} to BB, and time t3t_3 =4 s=4 \mathrm{~s} from point BB to highest point. Find the height of the middle point of AA and BB from the ground.
View answer and explanation
Correct option: A
60 m
Option A (60 m) is correct because it is the height of the midpoint of A and B calculated using kinematics and the given time intervals.
77 Medium
77 of 333
Analyzing +4 | -1
Assertion : A body is momentarily at rest when it reverses the direction.

Reason : A body cannot have acceleration if its velocity is zero at a given instant of time.
View answer and explanation
Correct option: C
Assertion is correct, reason is incorrect
Option C is correct because the average rotational kinetic energy of a diatomic molecule is kT, which is mv²/6, not mv²/4.
78 Medium
78 of 333
Analyzing +4 | -1
Statement-I: When a particle moves with constant velocity its average velocity, its instantaneous velocity and its speed are all equal in magnitude.
Statement-II: If the average velocity of a particle moving on a straight line is zero in a time interval, then it is possible that the instantaneous velocity is never zero in the interval
View answer and explanation
Correct option: D
Both statements are false
Option D (Both statements are false) is correct because Statement-I is false as speed is always positive, and Statement-II is false because if average velocity is zero, the instantaneous velocity must be zero at some point due to path reversal.
79 Medium
79 of 333
Application +4 | -1
A stone is dropped into a well in which the level of water is h\mathrm{h} below the top of the well. If v\mathrm{v} is velocity of sound, the time T\mathrm{T} after which the splash is heard is given by
View answer and explanation
Correct option: B
T=(2 h g)+hv\mathrm{T}=\sqrt{\left(\frac{2 \mathrm{~h}}{\mathrm{~g}}\right)}+\frac{\mathrm{h}}{\mathrm{v}}
Option B (T=(2 h g)+hv\mathrm{T}=\sqrt{\left(\frac{2 \mathrm{~h}}{\mathrm{~g}}\right)}+\frac{\mathrm{h}}{\mathrm{v}}) is correct because it represents the sum of the time taken for the stone to fall and the time taken for the sound to travel back up.
80 Medium
80 of 333
Understanding +4 | -1
A ball P is dropped vertically and another ball Q is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, then
View answer and explanation
Correct option: C
Both reach the ground at the same time
Option C (Both reach the ground at the same time) is correct because the vertical motion of both balls is governed by gravity and is independent of the horizontal motion.
81 Medium
81 of 333
Analyzing +4 | -1
Assertion (A) : Acceleration of a moving particle can change its direction without any change in direction of velocity.

Reason (R) : If the direction of change in velocity vector changes, the direction of acceleration vector also changes.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
82 Medium
82 of 333
Application +4 | -1
A drunkard is walking along a straight road. He takes five steps forward and three steps backward and so on. Each step is 1 m1 \mathrm{~m} long and takes 1 s1 \mathrm{~s}. There is a pit on the road 11 m11 \mathrm{~m} away from the starting point. The drunkard will fall into the pit after
View answer and explanation
Correct option: A
29 s29 \mathrm{~s}
Option A (29 s29 \mathrm{~s}) is correct because after 24 seconds the drunkard is 6m away, and then moves 5m forward in 5 seconds to fall into the 11m pit.
83 Medium
83 of 333
Analyzing +4 | -1
The incorrect statement(s) from the following is/are

I. A body having zero velocity will not necessarily have zero acceleration

II. A body having zero velocity will necessarily have zero acceleration

III. A body having uniform speed can have only uniform acceleration

IV. A body having non-uniform velocity will have zero acceleration
View answer and explanation
Correct option: A
II, III and IV
Option A (II, III and IV) is correct because statements II, III, and IV are incorrect regarding velocity and acceleration.
84 Medium
84 of 333
Understanding +4 | -1
A car covers a distance of S\mathrm{S} on a straight road in 3 hours and then returns to the initial point in four hours. Then, its average velocity is
View answer and explanation
Correct option: C
0
Option C is correct. Average velocity is defined as total displacement divided by total time. Since the car returns to its initial point, the total displacement is zero, making the average velocity zero.
85 Medium
85 of 333
Application +4 | -1
A projectile is fired vertically upwards with an initial velocity u. After an interval of T seconds a second projectile is fired vertically upwards, also with initial velocity is
View answer and explanation
Correct option: C
They meet at time t=ug+T2t = \frac{u}{g} + \frac{T}{2} and at a height u22g−gT28\frac{{{u^2}}}{{2g}} - \frac{{g{T^2}}}{8}
Option C (They meet at time t=ug+T2t = \frac{u}{g} + \frac{T}{2} and at a height u22g−gT28\frac{{{{u^2}}}}{{2g}} - \frac{{g{{T^2}}}}{8}) is correct because it is derived from equating the heights of the two projectiles at a given time.
86 Medium
86 of 333
Application +4 | -1
A car moving with constant acceleration crosses two points A and B with velocities 10 m/s and 20 m/s in its path. Find the velocity of the car midway between A and B (in m/s)
View answer and explanation
Correct option: D
5105\sqrt {10}
Option D (5105\sqrt {10}) is correct because it is calculated using the equation v2=u2+2asv^2 = u^2 + 2as to find the velocity at the midpoint.
87 Medium
87 of 333
Application +4 | -1
A point particle starting from rest has a velocity that increase linearly with time such that v=pt, where p=4ms−2p = 4\mathrm{m}{\mathrm{s}^{ - 2}}. The distance covered in the first 2 s will be
View answer and explanation
Correct option: C
8 m
Option C (8 m) is correct because the distance is calculated by integrating the velocity function v=pt with respect to time from 0 to 2 seconds.
88 Medium
88 of 333
Application +4 | -1
Find value of a if distance between the points (−9 cm,acm)(-9 \mathrm{~cm}, \mathrm{a} \mathrm{cm}) and (3 cm,3 cm(3 \mathrm{~cm}, 3 \mathrm{~cm} ) is 13 cm13 \mathrm{~cm}.
View answer and explanation
Correct option: A
8 cm8 \mathrm{~cm}
Option A (8 cm8 \mathrm{~cm}) is correct because it satisfies the distance formula between the two points.
89 Medium
89 of 333
Application +4 | -1
The coordinates of a moving particle at any time tt are given by, x=2t3x=2 t^3 and y=3t3y=3 t^3. Acceleration of the particle is given by
View answer and explanation
Correct option: B
t468t \sqrt{468}
Option B (t468t \sqrt{{468}}) is correct because it is the magnitude of the acceleration vector calculated from the given coordinates.
90 Medium
90 of 333
Application +4 | -1
An aeroplane files around a square field ABCD of each side 1000 km. Its speed along AB is 250  km  h−1250\;km\;{h^{ - 1}}, along DA 100  km  h−1100\;km\;{h^{ - 1}}. Its average speed (in km  h−1km\;{h^{ - 1}})) over the entire trip is
View answer and explanation
Correct option: D
190.5
Option D (190.5) is correct because the average speed is calculated by dividing the total distance by the total time taken for the entire trip.
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 61-90 of 333 questions

Your practice

0 / 333 answered · 333 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.