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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 211-240 of 285 questions

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211 Medium
211 of 285
Application +4 | -1
A train 175 m175 \mathrm{~m} long crosses a bridge which is 225 m225 \mathrm{~m} long in 80 seconds. What is the speed of the train?
View answer and explanation
Correct option: A
5 m/sec5 \mathrm{~m} / \mathrm{sec}
Option A (5 m/sec5 \mathrm{~m} / \mathrm{sec}) is correct because the speed is calculated by dividing the total distance (train length + bridge length) by the time taken.
212 Medium
212 of 285
Analyzing +4 | -1
The motion of a particle moving along yy-axis is represented as y=3(t−2)+5(t−2)2.y=3(t-2)+5(t-2)^2. Identify the correct statement.
View answer and explanation
Correct option: C
The particle is at the origin at t=2 st=2 \mathrm{~s}
Option C is correct. The particle is at the origin at t=2s because substituting t=2 into the equation gives y=0.
213 Medium
213 of 285
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.
214 Medium
214 of 285
Analyzing +4 | -1
Assertion (A) : A body is dropped from height hh and another body is thrown vertically upwards with a speed gh\sqrt{g h}. They meet at height h/2h / 2.

Reason (R) : The time taken by both the blocks in reaching the height h/2h / 2 is same.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
The correct answer is A because both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion. Both objects reach h/2 in the same time.
215 Medium
215 of 285
Application +4 | -1
The displacement of a particle starting from rest (at t=0) is given by s=6t2−t3.s = 6{t^2} - {t^3}.. The time in seconds at which the particle will attain zero velocity again, is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the velocity is zero when the derivative of the displacement function equals zero, which occurs at t=4 seconds.
216 Medium
216 of 285
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 represents the ratio of average velocity to maximum velocity, calculated using the distances and velocities during different phases of motion.
217 Medium
217 of 285
Application +4 | -1
Two trains 121 m121 \mathrm{~m} and 99 m99 \mathrm{~m} in lengths are running in opposite directions with velocities 40 km/h40 \mathrm{~km} / \mathrm{h} and 32 km/h32 \mathrm{~km} / \mathrm{h}. In what time they will completely cross each other?
View answer and explanation
Correct option: B
11 sec
Option B (11 sec) is correct because it correctly calculates the relative velocity and the total distance to find the time taken to cross each other.
218 Medium
218 of 285
Application +4 | -1
A person travels along a straight road for the first half time with a velocity v1v_1 and the second half time with a velocity v2v_2. Then the mean velocity vˉ\bar{v} is given by
View answer and explanation
Correct option: A
vˉ=v1+v22\bar{v}=\frac{v_1+v_2}{2}
Option A (vˉ=v1+v22\bar{v}=\frac{v_1+v_2}{2}) is correct because the average velocity is calculated as the arithmetic mean of the two velocities when the time intervals are equal.
219 Medium
219 of 285
Application +4 | -1
A bus starts from rest and moves along a straight road with constant acceleration ff until its speed is vv. The bus then continues with constant speed vv. When the bus starts, a car is at a distance bb behind the bus and is moving with constant speed uu in same direction as that of bus. Find the distance between the car and the bus at time tt after the bus started.

(Given 0<t<vf0 < t < \frac{v}{f} and u2<2fbu^2 < 2 f b).
View answer and explanation
Correct option: A
b−ut+ft22b-u t+\frac{f t^2}{2}
Option A (b−ut+ft22b-u t+\frac{f t^2}{2}) is correct because it represents the difference between the initial distance, the distance covered by the car, and the distance covered by the bus with constant acceleration.
220 Medium
220 of 285
Application +4 | -1
A body is moving with uniform velocity of 8 ms−18 \mathrm{~ms}^{-1}. When the body just crossed another body, the second one starts and moves with uniform acceleration of 4 ms−24 \mathrm{~ms}^{-2}. The time after which two bodies meet, will be
View answer and explanation
Correct option: B
4 s4 \mathrm{~s}
Option B (4 s4 \mathrm{~s}) is correct because the time is calculated by equating the displacements of both bodies.
221 Medium
221 of 285
Application +4 | -1
A train is moving with a velocity of 30 ms−130 \mathrm{~ms}^{-1}. When brakes are applied, it is found that the velocity reduced to 10 ms−110 \mathrm{~ms}^{-1} in 240 m240 \mathrm{~m}. When the velocity of the train is zero, the distance travelled is
View answer and explanation
Correct option: D
270 m270 \mathrm{~m}
Option D (270 m270 \mathrm{~m}) is correct because it is calculated by first finding the deceleration and then using it to find the total distance traveled.
222 Medium
222 of 285
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.
223 Medium
223 of 285
Application +4 | -1
A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball then rise to nearly a height
View answer and explanation
Correct option: D
0.64 h
Option D (0.64 h) is correct because the height reached after bouncing is proportional to the square of the velocity, which is 80% of the initial velocity.
224 Medium
224 of 285
Understanding +4 | -1
Two stones of equal masses are dropped from a rooftop of height h one after another. Their separation distance against time will
View answer and explanation
Correct option: B
Increase
Option B is correct because the separation distance increases with time due to the constant acceleration of gravity.
225 Medium
225 of 285
Analyzing +4 | -1
Assertion(A): Average velocity can't be zero in case of uniform acceleration.

Reason (R): For average velocity to be zero, a non zero velocity should not remain constant.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true) is correct because average velocity can be zero even with uniform acceleration (e.g., projectile motion), and the reason correctly states the condition for zero average velocity.
226 Medium
226 of 285
Analyzing +4 | -1
Assertion: In one-dimensional motion of an object, path length comes out to be equal to displacement.
Reason: Direction of object is changed two times.
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 in one-dimensional motion, path length equals displacement only if the direction doesn't change.
227 Medium
227 of 285
Analyzing +4 | -1
Assertion: An object may have varying speed without having varying velocity.
Reason: If the velocity is zero at an instant, the acceleration may not be zero at that instant.
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 the assertion is false, while the reason is true.
228 Medium
228 of 285
Understanding +4 | -1
Assertion: For a moving object the instantaneous velocity is calculated by the limiting value of its average velocity as the time interval approaches to zero.
Reason: The direction of the average velocity is same as the direction of displacement.
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 is correct because both the assertion and reason are true, but the reason doesn't explain why the instantaneous velocity is the limit of the average velocity.
229 Medium
229 of 285
Understanding +4 | -1
Assertion: For an object in uniform motion, the position-time graph is a straight line inclined to the time-axis.

Reason: In a uniform motion, the acceleration is zero.
View answer and explanation
Correct option: B
If both assertion and reason are true, but reason is not correct explanation of the assertion.
Option B is correct because both the assertion and the reason are true, but the reason (zero acceleration) doesn't directly explain why the position-time graph is a straight line; the constant velocity is the direct cause.
230 Medium
230 of 285
Application +4 | -1
A particle starts from the origin of coordinates at time t=0 and moves in the x-y plane with a constant acceleration&nbsp;α\alpha in the y-diretion. Its equation of motion is y=βx2y = \mathrm{\beta }{x^2}. Its velocity component in the x-direction is
View answer and explanation
Correct option: D
α2β\sqrt {\frac{\mathrm{\alpha }}{{2\mathrm{\beta }}}}
Option D (α2β\sqrt {\frac{{\alpha }}{{2\mathrm{\beta }}}}) is correct because the velocity component in the x-direction is constant and can be derived from the given equations of motion.
231 Medium
231 of 285
Analyzing +4 | -1
A particle moves in a straight line, so that after tt second the distance xx from a fixed point OO on the line is given by x=(t−2)2(t−5)x=(t-2)^2(t-5). Then
View answer and explanation
Correct option: C
the acceleration is negative when t<3 st<3 \mathrm{~s}.
Option C (the acceleration is negative when t<3 s) is correct because the second derivative of the position function is negative for t<3.
232 Medium
232 of 285
Application +4 | -1
A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for
View answer and explanation
Correct option: B
5s
Option B (5s) is correct because the time is calculated by equating the distance traveled in the last second to the distance traveled in the first three seconds.
233 Medium
233 of 285
Understanding +4 | -1
Assertion: An object can have an acceleration even if its speed is zero at a given instant.

Reason: An object was momentarily at rest when reversing its direction of travel.
View answer and explanation
Correct option: A
Both Assertion and Reason are true, and Reason is the correct explanation for Assertion
The correct answer is A (Both Assertion and Reason are true, and Reason is the correct explanation for Assertion) because an object can have zero speed momentarily while still experiencing acceleration, such as when reversing direction.
234 Medium
234 of 285
Application +4 | -1
A ball is thrown vertically upwards with a velocity of 25  ms−125\;m{s^{ - 1}} from the top of a tower of height 30 m. How long will it travel before it hits ground
View answer and explanation
Correct option: A
6s
Option A (6s) is correct because it is the positive root of the quadratic equation derived from the kinematic equation of motion.
235 Medium
235 of 285
Analyzing +4 | -1
A ball is dropped form the building of height 45  m45\,\,m simultaneously another ball is thrown up with a speed 40  m/s40\,\,m/s. The rate of change of relative speed of the ball is [ms−1]\left[ {m{s^{ - 1}}} \right]
View answer and explanation
Correct option: D
0
Option D (0) is correct because the relative acceleration between the two balls is zero, hence the rate of change of relative speed is zero.
236 Medium
236 of 285
Application +4 | -1
A ball is released from a height h\mathrm{h} above the ground. It takes a time T\mathrm{T} to reach the ground. Where is the ball at the time T/2\mathrm{T} / 2
View answer and explanation
Correct option: C
At a height 3 h/43 \mathrm{~h} / 4 from the ground
Option C (At a height 3 h/43 \mathrm{~h} / 4 from the ground) is correct because the distance fallen is proportional to the square of the time, so at half the time, the ball has fallen one-fourth of the total height.
237 Medium
237 of 285
Application +4 | -1
A stone dropped from a balloon which is at a height h, reaches the ground after t second. From the same balloon, if two stones are thrown, one upwards and the other downwards, with the same velocity u and they reach the ground after t1{t_1} and t2{t_2} second respectively, then
View answer and explanation
Correct option: C
t=t1t2t = \sqrt {{t_1}{t_2}}
Option C (t=t1t2t = \sqrt {{t_1}{t_2}}) is correct because it correctly relates the time taken for a stone dropped from a height h to the times taken when thrown upwards and downwards with the same velocity.
238 Medium
238 of 285
Analyzing +4 | -1
For a 1−D1-D motion, relation between position xx and time tt is t=αx2+βxt=\alpha x^2+\beta x. Find the relation between velocity vv and acceleration aa.
View answer and explanation
Correct option: C
a=−2αv3a=-2 \alpha v^3
Option C (a=−2αv3a=-2 \alpha v^3) is correct because it is derived by differentiating the given equation with respect to time and using the chain rule to relate acceleration and velocity.
239 Medium
239 of 285
Application +4 | -1
A body is fired vertically upward. At half the maximum height, the velocity of the body is 10m/s. The maximum height raised by the body is
View answer and explanation
Correct option: B
10 m
Option B (10 m) is correct because it is the maximum height calculated using kinematics equations, given the velocity at half the maximum height.
240 Medium
240 of 285
Application +4 | -1
The velocity vv of a particle as a function of its position (x)(x) is expressed as v=c1−c2xv=\sqrt{c_1-c_2 x}, where c1c_1 and c2c_2 are positive constants. The acceleration of the particle is
View answer and explanation
Correct option: B
−c22-\frac{c_2}{2}
Option B (−c22-\frac{c_2}{2}) is correct because the acceleration is found by using the chain rule to differentiate the velocity with respect to position and then multiplying by the velocity.
Revision notes and key takeaways

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