One-Dimensional Motion Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A stone is dropped from a height of \( 122.5 \, \text{m} \). 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, \( y = 122.5 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \).

Substitute: \( 122.5 = \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow 122.5 = 5 t^2 \Rightarrow t^2 = 24.5 \Rightarrow t = \sqrt{24.5} \approx 4.95 \, \text{s} \).

Rounded to one decimal, the time is \( 5 \, \text{s} \).

5 s
4 s
6 s
7 s
1

A runner starts from rest and accelerates at \( 1.5 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then maintains constant speed. What is the average speed over the first \( 10 \, \text{s} \)?

Acceleration: \( v = 1.5 \cdot 4 = 6 \, \text{m/s} \), \( x_1 = \frac{1}{2} \cdot 1.5 \cdot (4)^2 = 12 \, \text{m} \).

Constant speed for 6 s: \( x_2 = 6 \cdot 6 = 36 \, \text{m} \).

Total distance = \( 12 + 36 = 48 \, \text{m} \), total time = \( 10 \, \text{s} \).

Average speed = \( \frac{48}{10} = 4.8 \, \text{m/s} \).

4 m/s
5 m/s
4.8 m/s
6 m/s
3

A body accelerates uniformly from rest to \( 22 \, \text{m/s} \) over a distance of \( 121 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 22 \, \text{m/s} \), \( x = 121 \, \text{m} \).

Substitute: \( (22)^2 = 0 + 2 a (121) \Rightarrow 484 = 242 a \Rightarrow a = 2 \, \text{m/s}^2 \).

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

1 m/s²
3 m/s²
2 m/s²
4 m/s²
3

A car decelerates uniformly from \( 25 \, \text{m/s} \) to rest over a distance of \( 62.5 \, \text{m} \). What is the deceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v = 0 \), \( v_0 = 25 \, \text{m/s} \), \( x = 62.5 \, \text{m} \).

Substitute: \( 0 = (25)^2 + 2 a (62.5) \Rightarrow 0 = 625 + 125 a \Rightarrow a = -5 \, \text{m/s}^2 \).

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

2.5 m/s²
5 m/s²
7.5 m/s²
10 m/s²
2

A ball is thrown vertically upwards with a speed of \( 35 \, \text{m/s} \). What is the time taken to reach the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

At max height, \( v = 0 \). Use \( v = v_0 + a t \).

Here, \( v_0 = 35 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = 35 - 10 t \Rightarrow 10 t = 35 \Rightarrow t = 3.5 \, \text{s} \).

The time taken is \( 3.5 \, \text{s} \).

2.5 s
3 s
4 s
3.5 s
4

A stone is dropped from rest. What is the distance covered in the 7th second of its fall? (Take \( g = 10 \, \text{m/s}^2 \))

Distance in \( n \)-th second = \( g \cdot \frac{2n - 1}{2} \). For 7th second, \( n = 7 \).

Substitute: \( d = 10 \cdot \frac{2 \cdot 7 - 1}{2} = 10 \cdot \frac{13}{2} = 65 \, \text{m} \).

The distance covered is \( 65 \, \text{m} \).

55 m
60 m
70 m
65 m
4

A ball is thrown vertically upwards with a speed of \( 30 \, \text{m/s} \). What is the maximum height reached? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 a h \). At max height, \( v = 0 \), \( v_0 = 30 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (30)^2 + 2 (-10) h \Rightarrow 0 = 900 - 20 h \Rightarrow h = \frac{900}{20} = 45 \, \text{m} \).

The maximum height is \( 45 \, \text{m} \).

45 m
50 m
40 m
60 m
1

A ball is thrown upwards with a speed of \( 50 \, \text{m/s} \). How long does it take to return to the ground? (Take \( g = 10 \, \text{m/s}^2 \))

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

Total time = time up + time down = \( 5 \, \text{s} + 5 \, \text{s} = 10 \, \text{s} \).

The total time is \( 10 \, \text{s} \).

5 s
8 s
10 s
12 s
3

A runner moving at \( 5 \, \text{m/s} \) throws a ball horizontally at \( 15 \, \text{m/s} \) relative to himself. If it lands \( 18 \, \text{m} \) ahead of the throw point, what is the time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Ball’s speed relative to ground = \( 5 + 15 = 20 \, \text{m/s} \).

Relative distance: \( 20 t - 5 t = 18 \Rightarrow 15 t = 18 \Rightarrow t = 1.2 \, \text{s} \).

1 s
1.5 s
1.1 s
1.2 s
4

A projectile is launched vertically at \( 50 \, \text{m/s} \) from the ground. After \( 6 \, \text{s} \), a second projectile is launched at \( 40 \, \text{m/s} \). When do they meet? (Take \( g = 10 \, \text{m/s}^2 \))

First: \( y_1 = 50 t - 5 t^2 \), second: \( y_2 = 40 (t - 6) - 5 (t - 6)^2 \).

Meet when \( y_1 = y_2 \): \( 50 t - 5 t^2 = 40 (t - 6) - 5 (t^2 - 12 t + 36) \).

Simplify: \( 50 t - 5 t^2 = 40 t - 240 - 5 t^2 + 60 t - 180 \Rightarrow 50 t = 100 t - 420 \Rightarrow 50 t = 420 \Rightarrow t = 8.4 \, \text{s} \).

Time from second launch = \( 8.4 - 6 = 2.4 \, \text{s} \).

2 s
2.4 s
2.5 s
3 s
2

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