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  4. Motion In A Straight Line Chapter-Wise Test 3

Motion In A Straight Line Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A car moving at \( 50 \, \text{m/s} \) decelerates uniformly to rest over \( 125 \, \text{m} \). What is the time taken to stop?

Use \( v^2 = v_0^2 + 2 a x \) to find \( a \): \( 0 = (50)^2 + 2 a (125) \Rightarrow 0 = 2500 + 250 a \Rightarrow a = -10 \, \text{m/s}^2 \).

Then, \( v = v_0 + a t \): \( 0 = 50 - 10 t \Rightarrow t = 5 \, \text{s} \).

The time taken is \( 5 \, \text{s} \).

4 s
6 s
7 s
5 s
4

A stone is dropped from rest. What is the distance covered in the 2nd second of its fall? (Take \( g = 10 \, \text{m/s}^2 \))

Distance in \( n \)-th second = \( g \cdot \frac{2n - 1}{2} \). For 2nd second, \( n = 2 \).

Substitute: \( d = 10 \cdot \frac{2 \cdot 2 - 1}{2} = 10 \cdot \frac{3}{2} = 15 \, \text{m} \).

The distance covered is \( 15 \, \text{m} \).

10 m
20 m
5 m
15 m
4

A vehicle starts from rest and travels \( 50 \, \text{m} \) in \( 5 \, \text{s} \) with uniform acceleration. What is its final velocity?

Use \( x = v_0 t + \frac{1}{2} a t^2 \) to find \( a \): \( 50 = 0 \cdot 5 + \frac{1}{2} a (5)^2 \Rightarrow 50 = 12.5 a \Rightarrow a = 4 \, \text{m/s}^2 \).

Then, \( v = v_0 + a t = 0 + 4 \cdot 5 = 20 \, \text{m/s} \).

The final velocity is \( 20 \, \text{m/s} \).

10 m/s
20 m/s
15 m/s
25 m/s
2

A balloon descends at \( 6 \, \text{m/s} \) from \( 100 \, \text{m} \) height and releases a stone \( 2 \, \text{s} \) later. What is the time taken by the stone to hit the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Height at release: \( h = 100 - 6 \cdot 2 = 88 \, \text{m} \).

Stone’s initial velocity = \( -6 \, \text{m/s} \).

\( -88 = -6 t + \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow 5 t^2 - 6 t - 88 = 0 \).

Solve: \( t = \frac{6 \pm \sqrt{36 + 1760}}{10} = \frac{6 \pm 42.19}{10} \), \( t = 4.82 \, \text{s} \) (positive root).

4.5 s
4.82 s
5 s
4 s
2

A ball is thrown upwards at \( 24 \, \text{m/s} \) from a \( 50 \, \text{m} \) building. How far below the top of the building is it after \( 6 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement: \( y = 24 \cdot 6 - \frac{1}{2} \cdot 10 \cdot (6)^2 = 144 - 180 = -36 \, \text{m} \).

Distance below top = \( 36 \, \text{m} \).

36 m
30 m
40 m
35 m
1

A body starts from rest and accelerates uniformly at \( 5 \, \text{m/s}^2 \) for \( 3 \, \text{s} \). What is the distance covered?

Use \( x = v_0 t + \frac{1}{2} a t^2 \).

Here, \( v_0 = 0 \), \( a = 5 \, \text{m/s}^2 \), \( t = 3 \, \text{s} \).

Substitute: \( x = 0 \cdot 3 + \frac{1}{2} \cdot 5 \cdot (3)^2 = 0 + 2.5 \cdot 9 = 22.5 \, \text{m} \).

The distance covered is \( 22.5 \, \text{m} \).

15 m
30 m
22.5 m
45 m
3

A vehicle moving at \( 30 \, \text{m/s} \) decelerates uniformly to rest in \( 100 \, \text{m} \). What is the time taken to stop?

Use \( v^2 = v_0^2 + 2 a x \) to find \( a \): \( 0 = (30)^2 + 2 a (100) \Rightarrow 0 = 900 + 200 a \Rightarrow a = -4.5 \, \text{m/s}^2 \).

Then, \( v = v_0 + a t \): \( 0 = 30 - 4.5 t \Rightarrow t = \frac{30}{4.5} \approx 6.67 \, \text{s} \).

The time taken is approximately \( 6.7 \, \text{s} \).

5 s
6 s
7.5 s
6.7 s
4

A car decelerates uniformly from \( 16 \, \text{m/s} \) to rest over a distance of \( 32 \, \text{m} \). What is the deceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v = 0 \), \( v_0 = 16 \, \text{m/s} \), \( x = 32 \, \text{m} \).

Substitute: \( 0 = (16)^2 + 2 a (32) \Rightarrow 0 = 256 + 64 a \Rightarrow a = -4 \, \text{m/s}^2 \).

Magnitude of deceleration is \( 4 \, \text{m/s}^2 \).

2 m/s²
4 m/s²
6 m/s²
8 m/s²
2

A rocket is launched vertically at \( 60 \, \text{m/s} \) from the ground. After \( 5 \, \text{s} \), a second rocket is launched at \( 45 \, \text{m/s} \). How long after the second launch do they meet? (Take \( g = 10 \, \text{m/s}^2 \))

First: \( y_1 = 60 t - 5 t^2 \), second: \( y_2 = 45 (t - 5) - 5 (t - 5)^2 \).

Meet when \( y_1 = y_2 \): \( 60 t - 5 t^2 = 45 (t - 5) - 5 (t^2 - 10 t + 25) \).

Simplify: \( 60 t - 5 t^2 = 45 t - 225 - 5 t^2 + 50 t - 125 \Rightarrow 60 t = 95 t - 350 \Rightarrow 35 t = 350 \Rightarrow t = 10 \, \text{s} \).

Time from second launch = \( 10 - 5 = 5 \, \text{s} \).

4 s
5 s
4.5 s
6 s
2

A stone is thrown upwards at \( 15 \, \text{m/s} \) from a tower. It hits the ground \( 4 \, \text{s} \) later. What is the height of the tower? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( y = v_0 t + \frac{1}{2} a t^2 \). Downward is positive, so \( v_0 = -15 \, \text{m/s} \), \( a = 10 \, \text{m/s}^2 \).

Substitute: \( y = -15 \cdot 4 + \frac{1}{2} \cdot 10 \cdot (4)^2 = -60 + 80 = 20 \, \text{m} \).

Height = \( 20 \, \text{m} \).

15 m
25 m
20 m
30 m
2

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