Physics - Motion Straight Line - Quiz-1

A ball is thrown upward with a certain speed. It passes through the same point at 3 second and 7 second from the start. The maximum height achieved by the ball is:

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500 m
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250 m
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125 m
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450 m

For the following acceleration versus time graph the corresponding velocity versus displacement graph is

The displacement x of a particle moving in one dimension under the action of a constant force is related to time t by the equation $t = \sqrt{x} + 3$ , where x is in meters and t is in seconds. What is the displacement of the particle from t = 0 s to t = 6 s?

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0
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12 m
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6 m
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18 m

The acceleration a (in ${ms}^{- 2}$ ) of a body, starting from rest varies with time t(in s) following the equation a = 3t+4. The velocity of the body at time t = 2s will be :

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10 ${ms}^{- 1}$
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18 ${ms}^{- 1}$
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14 ${ms}^{- 1}$
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26 ${ms}^{- 1}$

A body thrown vertically so as to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is:

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$\sqrt{2} t$
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$(1 + \frac{1}{\sqrt{2}}) t$
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$\frac{3 t}{2}$
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$\frac{t}{\sqrt{2}}$

A stone falls freely from rest from a height h and it travels a distance $\frac{9 h}{25}$ in the last second. The value of h is:

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145 m
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100 m
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125 m
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200 ms

A point moves in a straight line under the retardation a $v^{2}$ . If the initial velocity is u, the distance covered in 't' seconds is-

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$aut$
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$\frac{1}{a} \ln (aut)$
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$\frac{1}{a} \ln (1 + aut)$
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$a \ln (aut)$

A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation of 2 $m / s^{2}$ . The ratio of time of ascent to the time of descent is: $[g = 10 m / s^{2}]$

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

A bullet loses $\frac{1}{20}$ of its velocity passing through a plank. The least number of planks required to stop the bullet is (All planks offers same retardation)

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10
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11
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12
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23

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4 t^{3} - 2 t)$ , where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin ?

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28 m/s²
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22 m/s²
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12 m/s²
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10 m/s²

The relation between time and distance is given by $t = α x^{2} + β x$ , where α and β are constants. The retardation, as calculated based on this equation, will be

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$2 α v^{3}$
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$2 β v^{3}$
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$2 α β v^{3}$
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$2 β^{2} v^{3}$

A point moves with uniform acceleration and v₁, v₂ and v₃ denote the average velocities in the three successive intervals of time t₁, t₂ and t₃. Which of the following relations is correct ?

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$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{2} + t_{3})$
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$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} + t_{2}) : (t_{2} + t_{3})$
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$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{1} - t_{3})$
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$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{2} - t_{3})$

The acceleration of a moving body can be found from:

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Area under the velocity-time graph
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Area under the distance-time graph
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Slope of the velocity-time graph
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Slope of the distance-time graph

The initial velocity of a particle is u (at t = 0) and the acceleration f is given by at. Which of the following relation is valid

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$v = u + a t^{2}$
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$v = u + a \frac{t^{2}}{2}$
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$v = u + a t$
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v = u

The initial velocity of the particle is 10 m/sec and its retardation is 2 m/sec². The distance moved by the particle in 5^th second of its motion is

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1 m
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19 m
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50 m
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75 m

A motor car moving with a uniform speed of 20 m/sec comes to stop on the application of brakes after travelling a distance of 10 m. Its acceleration is

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20 m/sec²
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–20 m/sec²
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–40 m/sec²
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+2 m/sec²

The velocity of a body moving with a uniform acceleration of 2 m/sec² is 10 m/sec. Its velocity after an interval of 4 sec is

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12 m/sec
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14 m/sec
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16 m/sec
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18 m/sec

A particle starting from rest moving with constant acceleration travels a distance x in first 2 seconds and a distance y in next two seconds, then

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y = x
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y = 2x
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y = 3x
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y = 4x

The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4 m/s². The distance covered by the body in the 5^th second of its motion is

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25 m
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35 m
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50 m
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85 m

The velocity of a body depends on time according to the equation $v = 20 + 0 .1 t^{2}$ . The body is undergoing

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Uniform acceleration
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Uniform retardation
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Non-uniform acceleration
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Zero acceleration

Which of the following four statements is false?

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A body can have zero velocity and still be accelerated.
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A body can have a constant velocity and still have a varying speed.
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A body can have a constant speed and still have a varying velocity.
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The direction of the velocity of a body can change when its acceleration is constant.

A particle moving with a uniform acceleration travels 24 m and 64 m in the first two consecutive intervals of 4 sec each. Its initial velocity is

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1 m/sec
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10 m/sec
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5 m/sec
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2 m/sec

The position of a particle moving in the XY plane at any time t is given by $x = (3 t^{2} - 6 t)$ metres. Select the correct statement about the moving particle from the following.

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The acceleration of the particle is zero at t = 0 second
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The velocity of the particle is zero at t = 0 second
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The velocity of the particle is zero at t = 1 second
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The velocity and acceleration of the particle are never zero

If body having initial velocity zero is moving with uniform acceleration 8 m/sec² , then the distance travelled by it in fifth second will be

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36 metres
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40 metres
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100 metres
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Zero

An alpha particle enters a hollow tube of 4 m length with an initial speed of 1 km/s. It is accelerated in the tube and comes out of it with a speed of 9 km/s. The time for which it remains inside the tube is

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$8 \times 10^{- 3} s$
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$80 \times 10^{- 3} s$
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$800 \times 10^{- 3} s$
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$8 \times 10^{- 4} s$

Two cars A and B are travelling in the same direction with velocities v₁ and v₂ $(v_{1} > v_{2})$ . When the car A is at a distance d behind car B, the driver of the car A applied the brake producing uniform retardation a. There will be no collision when-

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$d < \frac{{(v_{1} - v_{2})}^{2}}{2 a}$
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$d < \frac{v_{1}^{2} - v_{2}^{2}}{2 a}$
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$d > \frac{{(v_{1} - v_{2})}^{2}}{2 a}$
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$d > \frac{v_{1}^{2} - v_{2}^{2}}{2 a}$

A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 seconds on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is

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3 m/sec²
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–3 m/sec²
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0.3 m/sec²
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–0.3 m/sec²

A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the x-axis and $8 m / s^{2}$ along the y-axis. Its distance from the origin after 4 seconds will be

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56 m
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64 m
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80 m
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128 m

A car moving with a velocity of 10 m/s can be stopped by the application of a constant force F in a distance of 20 m. If the velocity of the car is 30 m/s, it can be stopped by this force in

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$\frac{20}{3} m$
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20 m
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60 m
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180 m

The displacement of a particle is given by $y = a + bt + {ct}^{2} - {dt}^{4}$ . The initial velocity and acceleration are, respectively:

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$b, - 4 d$
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$- b, 2 c$
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$b, 2 c$
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$2 c, - 4 d$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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