One-Dimensional Motion Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A bike moving at \( 126 \, \text{km/h} \) decelerates uniformly and stops in \( 7 \, \text{s} \). What is the magnitude of deceleration?

Convert speed: \( 126 \, \text{km/h} = 126 \cdot \frac{1000}{3600} = 35 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 35 \, \text{m/s} \), \( t = 7 \, \text{s} \).

Substitute: \( a = \frac{0 - 35}{7} = -5 \, \text{m/s}^2 \).

Magnitude of deceleration is \( 5 \, \text{m/s}^2 \).

5 m/s²
4 m/s²
6 m/s²
3 m/s²
1

A vehicle accelerates uniformly from rest at \( 1.2 \, \text{m/s}^2 \) for \( 10 \, \text{s} \). What is the final velocity?

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

Here, initial velocity \( v_0 = 0 \), acceleration \( a = 1.2 \, \text{m/s}^2 \), time \( t = 10 \, \text{s} \).

Substitute: \( v = 0 + 1.2 \cdot 10 = 12 \, \text{m/s} \).

The final velocity is \( 12 \, \text{m/s} \).

12 m/s
15 m/s
10 m/s
18 m/s
1

A train moving at \( 54 \, \text{km/h} \) accelerates at \( 1 \, \text{m/s}^2 \) for \( 10 \, \text{s} \), then decelerates at \( 2 \, \text{m/s}^2 \) to rest. What is the total distance covered?

Speed: \( 54 \, \text{km/h} = 15 \, \text{m/s} \).

Phase 1: \( v = 15 + 1 \cdot 10 = 25 \, \text{m/s} \), \( x_1 = 15 \cdot 10 + \frac{1}{2} \cdot 1 \cdot (10)^2 = 150 + 50 = 200 \, \text{m} \).

Phase 2: \( t = \frac{25}{2} = 12.5 \, \text{s} \), \( x_2 = 25 \cdot 12.5 - \frac{1}{2} \cdot 2 \cdot (12.5)^2 = 312.5 - 156.25 = 156.25 \, \text{m} \).

Total = \( 200 + 156.25 = 356.25 \, \text{m} \).

350 m
360 m
356.25 m
340 m
3

A person walks \( 6 \, \text{m} \) in \( 3 \, \text{s} \) and then walks back \( 2 \, \text{m} \) in \( 1 \, \text{s} \). What is the average speed?

Average speed = total distance / total time.

Total distance = \( 6 \, \text{m} + 2 \, \text{m} = 8 \, \text{m} \), total time = \( 3 \, \text{s} + 1 \, \text{s} = 4 \, \text{s} \).

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

1 m/s
1.5 m/s
2 m/s
2.5 m/s
3

A car moving at \( 90 \, \text{km/h} \) decelerates uniformly and stops in \( 12 \, \text{s} \). What is the magnitude of deceleration?

Convert speed: \( 90 \, \text{km/h} = 90 \cdot \frac{1000}{3600} = 25 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 25 \, \text{m/s} \), \( t = 12 \, \text{s} \).

Substitute: \( a = \frac{0 - 25}{12} = -2.083 \, \text{m/s}^2 \).

Magnitude of deceleration is approximately \( 2.08 \, \text{m/s}^2 \).

2.08 m/s²
1.5 m/s²
3 m/s²
2.5 m/s²
1

A body accelerates uniformly from rest to \( 28 \, \text{m/s} \) over a distance of \( 98 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 28 \, \text{m/s} \), \( x = 98 \, \text{m} \).

Substitute: \( (28)^2 = 0 + 2 a (98) \Rightarrow 784 = 196 a \Rightarrow a = 4 \, \text{m/s}^2 \).

The acceleration is \( 4 \, \text{m/s}^2 \).

2 m/s²
3 m/s²
4 m/s²
5 m/s²
3

A train moving at \( 108 \, \text{km/h} \) accelerates at \( 1.2 \, \text{m/s}^2 \) for \( 10 \, \text{s} \), then decelerates at \( 3 \, \text{m/s}^2 \) to rest. What is the total distance covered?

Speed: \( 108 \, \text{km/h} = 30 \, \text{m/s} \).

Phase 1: \( v = 30 + 1.2 \cdot 10 = 42 \, \text{m/s} \), \( x_1 = 30 \cdot 10 + \frac{1}{2} \cdot 1.2 \cdot (10)^2 = 300 + 60 = 360 \, \text{m} \).

Phase 2: \( t = \frac{42}{3} = 14 \, \text{s} \), \( x_2 = 42 \cdot 14 - \frac{1}{2} \cdot 3 \cdot (14)^2 = 588 - 294 = 294 \, \text{m} \).

Total = \( 360 + 294 = 654 \, \text{m} \).

650 m
660 m
654 m
640 m
3

A truck moves with a constant speed of \( 15 \, \text{m/s} \) for \( 40 \, \text{s} \). What is the distance covered?

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

Here, \( v = 15 \, \text{m/s} \), \( t = 40 \, \text{s} \).

Substitute: \( x = 15 \cdot 40 = 600 \, \text{m} \).

The distance covered is \( 600 \, \text{m} \).

600 m
400 m
800 m
500 m
1

A stone is thrown upwards with a speed of \( 20 \, \text{m/s} \). What is its velocity after \( 3 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v = v_0 + a t \). Here, \( v_0 = 20 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \), \( t = 3 \, \text{s} \).

Substitute: \( v = 20 - 10 \cdot 3 = 20 - 30 = -10 \, \text{m/s} \).

The velocity is \( -10 \, \text{m/s} \) (downward).

0 m/s
-10 m/s
10 m/s
-20 m/s
2

A rocket accelerates from rest at \( 7 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then moves at constant speed for \( 6 \, \text{s} \). What is the total distance covered?

Phase 1: \( v = 7 \cdot 4 = 28 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 7 \cdot (4)^2 = 56 \, \text{m} \).

Phase 2: \( x_2 = 28 \cdot 6 = 168 \, \text{m} \).

Total = \( 56 + 168 = 224 \, \text{m} \).

220 m
230 m
218 m
224 m
4

A train accelerates uniformly from \( 15 \, \text{m/s} \) to \( 35 \, \text{m/s} \) in \( 8 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{35 - 15}{8} = 2.5 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 15 \cdot 8 + \frac{1}{2} \cdot 2.5 \cdot (8)^2 = 120 + 80 = 200 \, \text{m} \).

The distance covered is \( 200 \, \text{m} \).

180 m
220 m
150 m
200 m
4

A stone falls freely from a height of \( 45 \, \text{m} \). How long does it take to reach the ground? (Take \( g = 10 \, \text{m/s}^2 \))

For free fall, use \( y = \frac{1}{2} g t^2 \).

Here, \( y = 45 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \).

Substitute: \( 45 = \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow 45 = 5 t^2 \Rightarrow t^2 = 9 \Rightarrow t = 3 \, \text{s} \).

The time taken is \( 3 \, \text{s} \).

3 s
2 s
4 s
5 s
1

A stone is dropped from rest. What is its velocity after falling \( 61.25 \, \text{m} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 10 \, \text{m/s}^2 \), \( y = 61.25 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 10 \cdot 61.25 = 1225 \Rightarrow v = \sqrt{1225} = 35 \, \text{m/s} \).

The velocity is \( 35 \, \text{m/s} \).

30 m/s
40 m/s
35 m/s
25 m/s
3

A car moving at \( 40 \, \text{m/s} \) decelerates uniformly to rest in \( 8 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{0 - 40}{8} = -5 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 40 \cdot 8 + \frac{1}{2} (-5) (8)^2 = 320 - 160 = 160 \, \text{m} \).

The distance covered is \( 160 \, \text{m} \).

120 m
200 m
160 m
80 m
3

A stone is dropped from rest. What is the distance covered in the 6th second of its fall? (Take \( g = 10 \, \text{m/s}^2 \))

Distance in \( n \)-th second = \( g \cdot \frac{2n - 1}{2} \). For 6th second, \( n = 6 \).

Substitute: \( d = 10 \cdot \frac{2 \cdot 6 - 1}{2} = 10 \cdot \frac{11}{2} = 55 \, \text{m} \).

The distance covered is \( 55 \, \text{m} \).

45 m
50 m
60 m
55 m
4

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