Physics - Motion Straight Line - Quiz-12

The position of a particle moving along the x-axis given by $x = (- 2 t^{3} + 3 t^{2} + 5) m$ . The acceleration of a particle at the instant its velocity becomes zero is

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12 m/s²
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-12 m/s²
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-6 m/s²
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Zero

A particle moves with velocity v₁ for time t₁ and v₂ for time t₂ along a straight line. The magnitude of its average acceleration is

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$\frac{v_{2} - v_{1}}{t_{1} - t_{2}}$
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$\frac{v_{2} - v_{1}}{t_{1} + t_{2}}$
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$\frac{v_{2} - v_{1}}{t_{2} - t_{1}}$
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$\frac{v_{1} + v_{2}}{t_{1} - t_{2}}$

A particle starts moving with acceleration 2 m/s². Distance travelled by it in 5^th half-second is-

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(1) 25 m
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(2) 25 m
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(3) 6.25 m
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(4) 30.25 m

The two ends of a train moving with constant acceleration pass a certain point with velocities u and 3u. The velocity with which the middle point of the train passes the same point is

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2u
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$\frac{3}{2} u$
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$\sqrt{5} u$
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$\sqrt{10} u$

A train starts from rest from a station with acceleration 0.2 m/s² on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s². If total time spent is half an hour, then the distance between two stations is- [Neglect length of train]

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216 km
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512 km
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728 km
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1296 km

A body is projected vertically in the upward direction from the surface of the earth. If the upward direction is taken as positive, then the acceleration of the body during its upward and downward journey is:

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Positive, negative
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Negative, negative
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Positive, positive
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Negative, positive

A particle starts moving from rest state along a straight line under the action of a constant force and travel distance x in first 5 seconds. The distance travelled by it in the next five seconds will be

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x
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2x
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3x
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4x

A body is projected vertically upward with a speed 40 m/s. The distance travelled by the body in the last second of the upward journey is [take g = 9.8 m/s² and neglect effect of air resistance]

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4.9 m
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9.8 m
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12.4 m
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19.6 m

A body is projected vertically upward with a speed 10 m/s and another at the same time with the same speed in the downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take g = 10 m/s²}

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Zero
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10 m/s²
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5 m/s²
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20 m/s²

A car travelling at a speed of 30 km/h is brought to rest at a distance of 8 m by applying brakes. If the same car is moving at a speed of 60 km/h, then it can be brought to rest with the same brakes in:

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64 m
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32 m
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16 m
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4 m

A particle is thrown with any velocity vertically upward, the distance travelled by the particle in the first second of its descent is:

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g
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$\frac{g}{2}$
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$\frac{g}{4}$
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Cannot be calculated

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance traveled by the body will be the same in

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1^st and 10^th second
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2^nd and 8^th second
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4^th and 6^th second
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Both (2) & (3)

A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of the second ball so that both hit the water at the same time is

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49 m/s
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55.5 m/s
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26.1 m/s
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9.8 m/s

A balloon starts from the ground from rest with an upward acceleration of 2 m/s². After 1 sec, a stone is dropped from it. The time taken by the stone to strike the ground is approximately

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0.3 s
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0.7 s
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1 s
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1.4 s

A boy throws balls into the air at a regular interval of 2 seconds. The next ball is thrown when the velocity of the first ball is zero. How high do the ball rise above his hand? [Take g = 9.8 m/ s²]

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4.9 m
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9.8 m
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19.6 m
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29.4 m

A ball projected from the ground vertically upward is at the same height at time t₁ and t₂. The speed of projection of ball is [Neglect the effect of air resistance]

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$g [t_{2} - t_{1}]$
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$\frac{g [t_{1} + t_{2}]}{2}$
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$\frac{g [t_{2} - t_{1}]}{2}$
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$g [t_{1} + t_{2}]$

Two balls are projected upward simultaneously with speeds of 40 m/s and 60 m/s. Relative position (x) of the second ball w.r.t. the first ball at time t = 5 sec will be [Neglect air resistance]

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20 m
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80 m
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100 m
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120 m

A ball is dropped from a height h above ground. Neglect the air resistance, its velocity (v) varies with its height y above the ground as

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$\sqrt{2 g (h - y)}$
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$\sqrt{2 g h}$
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$\sqrt{2 g y}$
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$\sqrt{2 g (h + y)}$

For a body moving with uniform acceleration along the straight line, the variation of its velocity (v) with position (x) is best represented by:

The position-time graph for a paricle moving along a straight line is shown in figure. The total distance travelled by it in time t=0 to t=10 s is

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Zero
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10 m
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20 m
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80 m

The position-time graph for a body moving along the straight line between O and A is shown in During its motion between O and A, how many times the body comes to rest?

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Zero
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1 time
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2 times
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3 times

Which one of the following graph for a body moving along a straight line is possible?

A body is projected vertically upward from ground. If we neglect the effect of air, then which one of te following is the best representation of variation of speed (v) with timw (t) ?

Which one of the following displacement-time graph represents two moving objects P and Q with zero relative velocity?

The displacement-time graph for two particles A and B is as follows. The ratio $\frac{v_{A}}{v_{B}}$ is

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1: 2
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1: $\sqrt{3}$
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$\sqrt{3}$ : 1
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1: 3

The velocity versus time graph of a body moving in a straight line is as shown in the figure below

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The distance covered by the body in 0 to 2 s is 8 m
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The acceleration of the body in 0 to 2 s is 4 ms^-2
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The acceleration of the body in 2 to 3 s is 4 ms^-2
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The distance moved by the body during 0 to 3 s is 6 m

The acceleration-time graph for a particle is given in the figure. If it starts motion at t=0, the distance travelled in 3 s will be

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4 m
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2 m
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0
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6 m

Figure shows the position of a particle moving on the x-axis as a function of time

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The particle has come to rest 4 times
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The velocity at t = 8 s is negative
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The velocity remains positive for t = 2 s to t = 6 s
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The particle moves with a constant velocity

A particle moves along x-axis in such a way that its x co-ordinate varies with time according to the equation x = 4 - 2t + t². The velocity of the particle will vary with time as-

The position (x) of a particle moving along x-axis vares with time (t) as shown in figure. The average acceleration of particle in time interval t=0 to t=8 s is

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3 m/s²
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- 5 m/s²
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- 4 m/s²
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2.5 m/s²

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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