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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 241-270 of 285 questions

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241 Medium
241 of 285
Application +4 | -1
The acceleration of a particle increasing linearly with time t is bt. The particle starts from the origin with an initial velocity v0{v_0}. The distance travelled by the particle in time t will be
View answer and explanation
Correct option: A
v0t+16bt3{v_0}t + \frac{1}{6}b{t^3}
Option A (v0t+16bt3{v_0}t + \frac{1}{6}b{t^3}) is correct because it is derived by integrating the acceleration twice with respect to time, considering the initial velocity.
242 Medium
242 of 285
Understanding +4 | -1
Which of the following is a one dimensional motion?
View answer and explanation
Correct option: D
Train running on a straight track
Option D (Train running on a straight track) is correct because one-dimensional motion involves movement along a single axis or straight line.
243 Medium
243 of 285
Understanding +4 | -1
A particle comes round a circle of radius 1 m once. The time taken by it is 10 sec. The average velocity of motion is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the average velocity is defined as the total displacement divided by the total time, and for a complete circle, the displacement is zero.
244 Medium
244 of 285
Analyzing +4 | -1
Assertion: A lift is ascending with decreasing speed means acceleration of lift is downwards.
Reason: A body always moves in the direction of its acceleration.
View answer and explanation
Correct option: C
If Assertion is true but Reason is false.
Option C (If Assertion is true but Reason is false) is correct because a lift ascending with decreasing speed has downward acceleration, but a body moves in the direction of its velocity, not necessarily its acceleration.
245 Medium
245 of 285
Application +4 | -1
An engine of a train moving with uniform acceleration passes the signal-post with velocity uu and the last compartment with velocity vv. The velocity with which middle point of the train passes the signal post is
View answer and explanation
Correct option: A
v2+u22\sqrt {\frac{{{v^2} + {u^2}}}{2}}
Option A (v2+u22\sqrt {\frac{{{v^2} + {u^2}}}{2}}) is correct because it is derived from applying the third equation of motion with uniform acceleration.
246 Medium
246 of 285
Application +4 | -1
A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is
View answer and explanation
Correct option: C
vg[1+1+2ghv2]\frac{v}{g}\left[ {1 + \sqrt {1 + \frac{{2gh}}{{{v^2}}}} } \right]
Option C (vg[1+1+2ghv2]\frac{v}{g}\left[ {1 + \sqrt {1 + \frac{{2gh}}{{{v^2}}}} } \right]) is correct because it is derived from solving the quadratic equation obtained by applying the equation of motion with constant acceleration.
247 Medium
247 of 285
Understanding +4 | -1
Assertion: Doubling the initial velocity, the stopping distance of moving object increases by a factor of 4 (for the same deceleration).
Reason: Stopping distance is proportional to the square of the initial velocity.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because the stopping distance is indeed proportional to the square of the initial velocity, which explains the assertion.
248 Medium
248 of 285
Understanding +4 | -1
Which of the following statement/s is/are correct?

Statement I: Only single value of distance is possible between two fixed points.

Statement II: Infinite values of displacements are possible between two fixed points.
View answer and explanation
Correct option: D
Neither I nor II
Option D is correct. Infinite distances are possible between two fixed points due to infinite possible paths, and only one displacement is possible.
249 Medium
249 of 285
Application +4 | -1
A body start from rest and moves with constant acceleration. The ratio of distance covered by the body in n thn\,{\text{th}} second to that covered in nn second is
View answer and explanation
Correct option: B
2n−1n2\frac{{2n - 1}}{{{n^2}}}
Option B (2n−1n2\frac{{2n - 1}}{{{n^2}}}) is correct because it correctly represents the ratio of distance covered in the nth second to that covered in n seconds for constant acceleration.
250 Medium
250 of 285
Understanding +4 | -1
A ball of mass mm and another ball of mass 2m2 m are dropped from equal heights. If the time taken by the balls to reach the ground are t1t_1 and t2t_2 respectively, then (neglect air drag)
View answer and explanation
Correct option: A
t1=t2t_1=t_2
Option A (t1=t2t_1=t_2) is correct because the time taken to fall from a height is independent of mass when air resistance is neglected.
251 Medium
251 of 285
Application +4 | -1
From the top of a tower of height 50m, a ball is thrown vertically upwards with a certain velocity. It hits the ground 10 s after it is thrown up. How much time does it take to cover a distance AB where A and B are two points 20m and 40m below the edge of the tower? (g=10ms−2\mathrm{g} = 10\mathrm{m}{\mathrm{s}^{ - 2}})
View answer and explanation
Correct option: D
0.4 s
Option D (0.4 s) is correct because it is the time difference between reaching 20m and 40m below the tower, calculated using kinematic equations.
252 Medium
252 of 285
Analyzing +4 | -1
A particle covers 4m, 5m, 6m, and 7m, in 3rd, 4th, 5th and 6th second respectively. The particle starts
View answer and explanation
Correct option: A
With an initial non-zero velocity and moves with uniform acceleration
Option A is correct because the particle's motion indicates a non-zero initial velocity and uniform acceleration, as derived from the given distances covered in successive seconds.
253 Medium
253 of 285
Application +4 | -1
The position of a particle moving along xx-axis varies with time tt as 4x+t2=44 x+t^2=4. At what position, the particle is at rest?
View answer and explanation
Correct option: C
x=1x=1
Option C (x=1x=1) is correct because the particle is at rest when its velocity is zero, which occurs at x=1.
254 Medium
254 of 285
Application +4 | -1
A particle moves in a straight line so that its displacement x metre at time t second is given by
t=x2−1t = \sqrt {{x^2} - 1} Its acceleration in ms−2  \mathrm{m}{\mathrm{s}^{ - 2}}\mathrm{\;} at time t second is
View answer and explanation
Correct option: A
1x3\frac{1}{{{x^3}}}
Option A (1x3\frac{1}{{{x^3}}}) is correct because it is derived from differentiating the given equation twice with respect to time to find the acceleration.
255 Medium
255 of 285
Application +4 | -1
A body is projected with a velocity 60 m/s vertically upwards the distance travelled in last second of its motion [g=10  m/s2][g = 10\,\,m\mathrm{/}{s^2}]
View answer and explanation
Correct option: B
55
Option B (55) is correct because the distance travelled in the last second of upward motion is equivalent to the distance travelled in the first second of downward motion.
256 Medium
256 of 285
Analyzing +4 | -1
The motion of a particle along a straight line is described by the equation: x=8+12t−t3x=8+12 t-t^3, where x{x} is in meter and t{t} in second.

(i) the initial velocity of particle is 12 m/s12 \mathrm{~m} / \mathrm{s}

(ii) the retardation of particle when velocity is zero is 12 m/s212 \mathrm{~m} / \mathrm{s}^2

(iii) when acceleration is zero, displacement is 8 m8 \mathrm{~m}

(iv) the maximum velocity of particle is 12 m/s12 \mathrm{~m} / \mathrm{s}
View answer and explanation
Correct option: D
All options are correct
Option D (All options are correct) is correct because all the statements about the particle's motion derived from the given equation are true.
257 Medium
257 of 285
Understanding +4 | -1
Assertion: The magnitude of average velocity of an object over an interval of time is either smaller than or equal to the average speed of the object over the same interval.
Reason: Path length (distance) is either equal to or greater than the magnitude of displacement.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the average velocity's magnitude is less than or equal to the average speed, and this is because the path length is greater than or equal to the displacement's magnitude.
258 Medium
258 of 285
Application +4 | -1
A particle starts from rest and moving with constant acceleration. Let Sn{S}_{{n}} and Sn+1{S}_{{n}+1} be the distance travelled by the particle in the time intervals t=n−1{t}={n}-1 to t=n{t}={n} and t=n{t}={n} to t=n+1{t}={n}+1 respectively. The ratio SnSn+1\frac{S_n}{S_{n+1}} is :
View answer and explanation
Correct option: B
2n−12n+1\frac{2 n-1}{2 n+1}
Option B (2n−12n+1\frac{2 n-1}{2 n+1}) is correct because it represents the ratio of distances traveled in consecutive time intervals for an object starting from rest with constant acceleration.
259 Medium
259 of 285
Understanding +4 | -1
A man is sitting in a train which is moving with a velocity of 60 km/hour60 {~km} / {h} o u r. His speed with respect to the train is:
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the man is sitting still relative to the train, so his speed relative to the train is zero.
260 Medium
260 of 285
Application +4 | -1
Two ends of a train moving at constant acceleration pass a point with speed of 12 m/s12 \mathrm{~m} / \mathrm{s} and 24 m/s24 \mathrm{~m} / \mathrm{s} respectively. The speed with which middle point of the train passes the same point is (approximately)
View answer and explanation
Correct option: B
19 m/s19 \mathrm{~m} / \mathrm{s}
Option B (19 m/s19 \mathrm{~m} / \mathrm{s}) is correct because the speed at the midpoint is calculated using the equation vm=u2+v22v_m = \sqrt{\frac{u^2+v^2}{2}}.
261 Medium
261 of 285
Understanding +4 | -1
Which of the following statements is not correct?
View answer and explanation
Correct option: C
The sign of acceleration tells us whether the particles speed is increasing or decreasing
Option C is correct because the sign of acceleration only indicates the direction of the acceleration vector, not whether the speed is increasing or decreasing.
262 Medium
262 of 285
Understanding +4 | -1
Two spheres of same size, one of mas 2 kg and another of mass 4 kg, are dropped simultaneously from the top of Qutub Minar (height =72m). When they are 1 m above the ground, the two spheres have the same
View answer and explanation
Correct option: D
Acceleration
Option D is correct. Acceleration due to gravity is independent of the mass of the object. Therefore, both spheres will have the same acceleration.
263 Medium
263 of 285
Application +4 | -1
A particle moving along a straight line has a velocity v  ms−1v\;\mathrm{m}{\mathrm{s}^{ - 1}}, when it cleared a distance of y metre. These two are connected by the relation v=49+yv = \sqrt {49 + y}. When its velocity is 1 ms−1,\mathrm{m}{\mathrm{s}^{ - 1}},, its acceleration (in ms−2\mathrm{m}{\mathrm{s}^{ - 2}})is
View answer and explanation
Correct option: D
0.50.5
Option D (0.50.5) is correct because the acceleration is calculated using a=vdvdya = v\frac{dv}{dy} and substituting the given velocity.
264 Medium
264 of 285
Application +4 | -1
A stone is thrown vertically upward with a speed of 28 m/s28 \mathrm{~m} / \mathrm{s}. (a) Find the maximum height reached by the stone. (b) Find its velocity one second before it reaches the maximum height?
View answer and explanation
Correct option: D
40 m; 9.87 m/s
Option D (40 m; 9.87 m/s) is correct because it calculates the maximum height using kinematics and the velocity one second before reaching that height.
265 Medium
265 of 285
Application +4 | -1
A body of mass 2kg2kg has an initial velocity of 3m/s3m/s along x−axisx - axis and it is subjected to an acceleration of 2 m/s22\,m/{s^2} in Y-direction. The distance of the body from origin after 4s4s will be
View answer and explanation
Correct option: C
20 m20\,m
Option C (20 m20\,m) is correct because it is the magnitude of the displacement vector after 4 seconds, calculated using kinematic equations.
266 Medium
266 of 285
Application +4 | -1
Two bodies of different masses are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is
View answer and explanation
Correct option: C
45\frac{4}{5}
Option C (45\frac{4}{5}) is correct because the time taken to reach the ground is proportional to the square root of the height, and the ratio is calculated accordingly.
267 Medium
267 of 285
Analyzing +4 | -1
Select the incorrect statements from the following
S1 : Average velocity is path length divided by time interval
S2 : In general, speed is greater than the magnitude of the velocity
S3 : A particle moving in a given direction with a non-zero velocity can have zero speed
S4 : The magnitude of average velocity is the average speed
View answer and explanation
Correct option: C
S1, S3 and S4
Option C (S1, S3 and S4) is correct because average velocity is displacement divided by time, a particle with non-zero velocity cannot have zero speed, and the magnitude of average velocity is not always the average speed.
268 Medium
268 of 285
Application +4 | -1
A body is moving from rest under constant acceleration and let S1{S_1} be the displacement in the first (P-1) sec and S2{S_2} be the displacement in the first P sec. The displacement in (P2−P+1)th{\left( {{P^2} - P + 1} \right)^{th}} see will be
View answer and explanation
Correct option: A
S1+S2{S_1} + {S_2}
Option A (S1+S2{S_1} + {S_2}) is correct because the displacement in the (P2−P+1)th{\left( {{P^2} - P + 1} \right)^{th}} second is the sum of the displacements in the first (P-1) sec and the first P sec.
269 Medium
269 of 285
Application +4 | -1
A driver takes 0.20 s0.20 \mathrm{~s} to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km/h54 \mathrm{~km} / \mathrm{h} and the brakes cause a deceleration of 5.0 m/s25.0 \mathrm{~m} / \mathrm{s}^2, find the distance travelled by the car after he sees the need to put the brakes on.
View answer and explanation
Correct option: A
25.5 m
Option A (25.5 m) is correct because it accounts for both the distance traveled during the reaction time and the distance traveled while decelerating.
270 Medium
270 of 285
Application +4 | -1
The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4  m/s24\;m/{s^2}. The distance covered by the body in the   5th\;{5^{th}} second of its motion is
View answer and explanation
Correct option: A
25 m
Option A (25 m) is correct because it is calculated using the formula for the distance covered in the nth second of motion.
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