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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

1 m

19 m

50 m

75 m

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Explanation

$S_{n} = u - \frac{a}{2} (2 n - 1) = 10 - \frac{2}{2} (2 \times 5 - 1) = 1 m e t e r$

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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

20 m/sec²

–20 m/sec²

–40 m/sec²

+2 m/sec²

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Explanation

From $v^{2} = u^{2} + 2 a S$ ⇒ $0 = u^{2} + 2 a S$

⇒ $a = \frac{- u^{2}}{2 S} = \frac{- {(20)}^{2}}{2 \times 10} = - 20 m / s^{2}$

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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

12 m/sec

14 m/sec

16 m/sec

18 m/sec

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Explanation

$v = u + a t = 10 + 2 \times 4 = 18 m / \sec$

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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

y = x

y = 2x

y = 3x

y = 4x

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Explanation

$Dist . covered in 1^{st} 2 \sec : ‐ x = o + \frac{1}{2} a \times {(2)}^{2} = 2 a Dist . covered in 4 \sec : ‐ d = o + \frac{1}{2} \times a \times {(4)}^{2} = 8 a \Rightarrow y = d - x = 8 a - 2 a = 6 a \Rightarrow \frac{x}{y} = \frac{2 a}{6 a} \Rightarrow y = 3 x$

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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

25 m

35 m

50 m

85 m

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Explanation

$S_{n} = u + \frac{a}{2} [2 n - 1]$

$S_{5^{t h}} = 7 + \frac{4}{2} [2 \times 5 - 1] = 7 + 18 = 25 m$ .

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

Uniform acceleration

Uniform retardation

Non-uniform acceleration

Zero acceleration

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Explanation

Acceleration $a = \frac{d v}{d t} = 0 .1 \times 2 t = 0 .2 t$

Which is time dependent i.e. non-uniform acceleration.

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Which of the following four statements is false ?

A body can have zero velocity and still be accelerated

A body can have a constant velocity and still have a varying speed

A body can have a constant speed and still have a varying velocity

The direction of the velocity of a body can change when its acceleration is constant

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Explanation

Constant velocity means constant speed as well as same direction throughout.

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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

1 m/sec

10 m/sec

5 m/sec

2 m/sec

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Explanation

Distance travelled in 4 sec

$24 = 4 u + \frac{1}{2} a \times 16$ …(i)

Distance travelled in total 8 sec

$88 = 8 u + \frac{1}{2} a \times 64$ …(ii)

After solving (i) and (ii), we get u = 1 m/s.

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

The acceleration of the particle is zero at t = 0 second

The velocity of the particle is zero at t = 0 second

The velocity of the particle is zero at t = 1 second

The velocity and acceleration of the particle are never zero

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Explanation

$v_{x} = \frac{d x}{d t} = \frac{d}{d t} (3 t^{2} - 6 t) = 6 t - 6$ . At $t = 1, v_{x} = 0$

$v_{y} = \frac{d y}{d t} = \frac{d}{d t} (t^{2} - 2 t) = 2 t - 2$ . At $t = 1, v_{y} = 0$

Hence $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 0$

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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

36 metres

40 metres

100 metres

Zero

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Explanation

Distance travelled in $n^{t h}$ second $= u + \frac{a}{2} (2 n - 1)$

Distance travelled in $5^{t h}$ second $= 0 + \frac{8}{2} (2 \times 5 - 1)$ = 36m

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