One-Dimensional Motion Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A truck moving at \( 144 \, \text{km/h} \) decelerates uniformly and stops in \( 12 \, \text{s} \). What is the magnitude of deceleration?

Convert speed: \( 144 \, \text{km/h} = 144 \cdot \frac{1000}{3600} = 40 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 40 \, \text{m/s} \), \( t = 12 \, \text{s} \).

Substitute: \( a = \frac{0 - 40}{12} = -3.33 \, \text{m/s}^2 \).

Magnitude of deceleration is approximately \( 3.33 \, \text{m/s}^2 \).

3.33 m/s²
4 m/s²
2.5 m/s²
5 m/s²
1

A car accelerates uniformly from \( 5 \, \text{m/s} \) to \( 25 \, \text{m/s} \) in \( 10 \, \text{s} \). What is the distance traveled?

Find \( a = \frac{v - v_0}{t} = \frac{25 - 5}{10} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 5 \cdot 10 + \frac{1}{2} \cdot 2 \cdot (10)^2 = 50 + 100 = 150 \, \text{m} \).

The distance traveled is \( 150 \, \text{m} \).

100 m
200 m
150 m
75 m
3

A cyclist moves at a constant speed of \( 12 \, \text{m/s} \) for \( 30 \, \text{s} \). What is the distance covered?

For constant speed, distance \( x = v t \).

Here, \( v = 12 \, \text{m/s} \), \( t = 30 \, \text{s} \).

Substitute: \( x = 12 \cdot 30 = 360 \, \text{m} \).

The distance covered is \( 360 \, \text{m} \).

300 m
320 m
400 m
360 m
4

A car accelerates uniformly from rest at \( 4 \, \text{m/s}^2 \) for \( 5 \, \text{s} \). What is the final velocity?

Use the equation \( v = v_0 + a t \).

Here, initial velocity \( v_0 = 0 \), acceleration \( a = 4 \, \text{m/s}^2 \), time \( t = 5 \, \text{s} \).

Substitute: \( v = 0 + 4 \cdot 5 = 20 \, \text{m/s} \).

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

20 m/s
25 m/s
15 m/s
10 m/s
1

A man walks \( 4 \, \text{km} \) in \( 1 \, \text{hour} \) and then returns to the starting point in \( 1 \, \text{hour} \). What is his average speed?

Average speed = total distance / total time.

Total distance = \( 4 \, \text{km} + 4 \, \text{km} = 8 \, \text{km} \), total time = \( 1 \, \text{h} + 1 \, \text{h} = 2 \, \text{h} \).

Average speed = \( \frac{8}{2} = 4 \, \text{km/h} \).

2 km/h
4 km/h
0 km/h
8 km/h
2

A cyclist moves at a constant speed of \( 6 \, \text{m/s} \) for \( 50 \, \text{s} \). What is the distance covered?

For constant speed, distance \( x = v t \).

Here, \( v = 6 \, \text{m/s} \), \( t = 50 \, \text{s} \).

Substitute: \( x = 6 \cdot 50 = 300 \, \text{m} \).

The distance covered is \( 300 \, \text{m} \).

250 m
350 m
200 m
300 m
4

A car accelerates from \( 8 \, \text{m/s} \) at \( 2 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then moves at constant speed for \( 3 \, \text{s} \). What is the average speed over the entire motion?

Phase 1: \( v = 8 + 2 \cdot 5 = 18 \, \text{m/s} \), \( x_1 = 8 \cdot 5 + \frac{1}{2} \cdot 2 \cdot (5)^2 = 40 + 25 = 65 \, \text{m} \).

Phase 2: \( x_2 = 18 \cdot 3 = 54 \, \text{m} \).

Total distance = \( 65 + 54 = 119 \, \text{m} \), time = \( 5 + 3 = 8 \, \text{s} \).

Average speed = \( \frac{119}{8} = 14.875 \, \text{m/s} \).

14 m/s
15 m/s
14.875 m/s
16 m/s
3

A ball is thrown vertically upwards with \( 30 \, \text{m/s} \). How long does it take to return to the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Time to reach max height: \( v = v_0 + a t \), \( 0 = 30 - 10 t \Rightarrow t = 3 \, \text{s} \).

Total time = time up + time down = \( 3 \, \text{s} + 3 \, \text{s} = 6 \, \text{s} \).

The total time is \( 6 \, \text{s} \).

3 s
4 s
5 s
6 s
4

A train accelerates uniformly from \( 20 \, \text{m/s} \) to \( 40 \, \text{m/s} \) in \( 10 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{40 - 20}{10} = 2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 20 \cdot 10 + \frac{1}{2} \cdot 2 \cdot (10)^2 = 200 + 100 = 300 \, \text{m} \).

The distance covered is \( 300 \, \text{m} \).

250 m
350 m
200 m
300 m
4

A stone is dropped from rest. What is the distance covered in the 5th second of its fall? (Take \( g = 10 \, \text{m/s}^2 \))

Distance in \( n \)-th second = \( g \cdot \frac{2n - 1}{2} \). For 5th second, \( n = 5 \).

Substitute: \( d = 10 \cdot \frac{2 \cdot 5 - 1}{2} = 10 \cdot \frac{9}{2} = 45 \, \text{m} \).

The distance covered is \( 45 \, \text{m} \).

35 m
40 m
50 m
45 m
4

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