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JEE • PHYSICS • MOTION IN A STRAIGHT LINE

Motion in a straight line practice · Mixed difficulty

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Showing 301-330 of 333 questions

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301 Medium
301 of 333
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.
302 Medium
302 of 333
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}}.
303 Hard
303 of 333
Analyzing +4 | -1
A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t1{t_1} and t2{t_2} seconds 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 represents the geometric mean of the times when the stones are thrown upwards and downwards, derived from kinematic equations.
304 Medium
304 of 333
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.
305 Medium
305 of 333
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.
306 Medium
306 of 333
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.
307 Medium
307 of 333
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.
308 Medium
308 of 333
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.
309 Medium
309 of 333
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.
310 Medium
310 of 333
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.
311 Medium
311 of 333
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.
312 Medium
312 of 333
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.
313 Medium
313 of 333
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.
314 Medium
314 of 333
Analyzing +4 | -1
Car AA is moving with a speed of 36 kmh−136 \mathrm{~kmh}^{-1} on a two lane road. Two cars BB and CC, each moving with a speed of 54 kmh−154 \mathrm{~kmh}^{-1} in opposite directions on the other lane are approaching car A. At certain instant of time, when the distance AB=AC=1 kmA B=A C=1 \mathrm{~km}, the driver of car BB decides to overtake AA before CC does. What must be the minimum acceleration of car BB, so as to avoid an accident ?
View answer and explanation
Correct option: A
1 ms−21 \mathrm{~ms}^{-2}
Option A (1 ms−21 \mathrm{~ms}^{-2}) is correct because it is the minimum acceleration required for car B to overtake car A before car C reaches car A.
315 Medium
315 of 333
Analyzing +4 | -1
A particle moves along a straight line such that its position x at any time t is x=6t2−t3.x = 6{t^2} - {t^3}.. Where x in metre and t is in second, then
View answer and explanation
Correct option: D
Both (1) and (2) are correct
Option D (Both (1) and (2) are correct) is correct because at t=0, the acceleration is 12 m/s², and the x-t curve has a maximum at t=4 s.
316 Medium
316 of 333
Understanding +4 | -1
Two bodies one held 100 cm100 \mathrm{~cm} directly above the other are released simultaneously from the large height and both fall freely under the gravity. The separation between the bodies just after 2 s2 \mathrm{~s} will be
View answer and explanation
Correct option: B
100 cm100 \mathrm{~cm}
Option B is correct because the separation between the two bodies remains constant at 100 cm as they fall under gravity.
317 Easy
317 of 333
Understanding +4 | -1
The velocity of a body at a given instant of time during motion is:
View answer and explanation
Correct option: C
Instantaneous velocity =lim⁡Δt→0Δs⃗Δt=\lim \limits_{{\Delta t \rightarrow 0}} \frac{\Delta \vec{s}}{\Delta t}
Option C is correct. Instantaneous velocity is defined as the limit of the average velocity as the time interval approaches zero.
318 Medium
318 of 333
Application +4 | -1
Two trains travelling on the same track are approaching each other with equal speeds of 40ms−140\mathrm{m}{\mathrm{s}^{ - 1}}. The drivers of the trains begin to decelerate simultaneously when they are just 2 km apart. If the decelerations are both uniform and equal, then the value of deceleration to barely avoid collision should be
View answer and explanation
Correct option: A
0.8ms−20.8\mathrm{m}{\mathrm{s}^{ - 2}}
Option A (0.8ms−20.8\mathrm{m}{\mathrm{s}^{ - 2}}) is correct because it is the deceleration required to avoid collision, calculated using kinematics.
319 Hard
319 of 333
Analyzing +4 | -1
If position vector of a particle at time tt is r⃗=3ti^+(8−4t)j^ m\vec{r}=3 t \hat{i}+(8-4 t) \hat{j} \mathrm{~m}, then
View answer and explanation
Correct option: A
Particle is moving with constant velocity in straight line
The correct answer is A because the velocity vector is constant, indicating motion in a straight line with constant velocity.
320 Medium
320 of 333
Analyzing +4 | -1
Which of the following 4 statements is false
View answer and explanation
Correct option: B
A body can have a constant velocity and still have a varying speed
Option B is correct. A body cannot have a constant velocity and still have a varying speed because constant velocity implies both constant speed and constant direction.
321 Medium
321 of 333
Application +4 | -1
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 s. Its average velocity is
View answer and explanation
Correct option: C
2 m/s2\,{\text{m/s}}
Option C (2 m/s) is correct because average velocity is calculated by dividing the displacement by the time taken, and in half a revolution, displacement is 2R.
322 Medium
322 of 333
Application +4 | -1
An object is moving with uniform acceleration. Its velocity after 5 seconds is 25 ms−125 \mathrm{~ms}^{-1} and after 8 seconds, it is 34 ms−1.34 \mathrm{~ms}^{-1}. Find the distance travelled by the object in 12th second.
View answer and explanation
Correct option: A
44.5 m44.5 \mathrm{~m}
Option A (44.5 m) is correct because the distance travelled in the 12th second is calculated using the kinematic equations of motion with uniform acceleration.
323 Medium
323 of 333
Application +4 | -1
A body is dropped from the top of a tower clears (7/16)th{\left( {7/16} \right)^{\mathrm{th}}} of total height of the tower in its last second of flight. The height of the tower is
View answer and explanation
Correct option: C
78.4 m
Option C (78.4 m) is correct because it is calculated by using the kinematic equation of motion and solving for the height of the tower.
324 Medium
324 of 333
Analyzing +4 | -1
Assertion: For a particle, acceleration-time graph gives the velocity.
Reason: Rate of change of velocity is acceleration.
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 while both statements are true, the reason doesn't explain why the area under the acceleration-time graph gives the change in velocity.
325 Easy
325 of 333
Understanding +4 | -1
A particle covers equal distance around a circular path, in equal intervals of time. Which of the following quantities connected with the motion of the particle remains constant with time?
View answer and explanation
Correct option: C
Speed
Option C is correct because covering equal distances in equal time intervals implies constant speed.
326 Hard
326 of 333
Application +4 | -1
A ball dropped from a point PP crosses a point QQ in tt seconds. The time taken by it to travel from Q to R, if PQ=QRPQ = QR
View answer and explanation
Correct option: D
(2−1)t\left( {\sqrt 2 - 1} \right)t
Option D ((2−1)t\left( {\sqrt 2 - 1} \right)t) is correct because it is the time taken to travel from Q to R, calculated using kinematic equations for constant acceleration.
327 Medium
327 of 333
Application +4 | -1
Two cars A and B at rest at same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with constant acceleration of 4m/s24m/{s^2}, then B will catch A after how much time
View answer and explanation
Correct option: B
20 sec
Option B (20 sec) is correct because the time is calculated by equating the distances traveled by both cars when B catches A.
328 Medium
328 of 333
Analyzing +4 | -1
I. Particle-1 is dropped from a tower and particle-2 is projected horizontal from the same tower, then both the particles reach the ground simultaneously.

II. Both the particles strike the ground with different speeds.

Which of the following statement(s) is/are correct?
View answer and explanation
Correct option: C
Both I and II
Option C is correct. Both particles reach the ground simultaneously because vertical motion is independent of horizontal motion, and they strike with different speeds due to the horizontal velocity component of particle-2.
329 Medium
329 of 333
Application +4 | -1
A particle moves along a straight line such that its displacement at any time t is given by S=t3−6t2+3t+4S = {t^3} - 6{t^2} + 3t + 4 metres. The velocity when the acceleration is zero is
View answer and explanation
Correct option: B
−9ms−1- 9m{s^{ - 1}}
Option B (-9 m/s⁻¹) is correct because it's the velocity calculated at the time when the acceleration is zero, found by differentiating the displacement function twice.
330 Medium
330 of 333
Application +4 | -1
A moving car possesses average velocities of 5 m/s, 10 m/s and 15 m/s. In the first, second and third seconds respectively. Then the total distance covered by the car in these 3 seconds is
View answer and explanation
Correct option: B
30 m
Option B (30 m) is correct because it's the sum of the distances covered in each second, calculated using the average velocities.
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