One-Dimensional Motion Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A ball is thrown upwards with a speed of \( 32 \, \text{m/s} \). 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, \( v = 0 \), \( v_0 = 32 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (32)^2 + 2 (-10) h \Rightarrow 0 = 1024 - 20 h \Rightarrow h = \frac{1024}{20} = 51.2 \, \text{m} \).

The maximum height is \( 51.2 \, \text{m} \).

40 m
48 m
51.2 m
64 m
3

A ball is thrown vertically upwards with a speed of \( 16 \, \text{m/s} \). 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, \( v = 0 \), \( v_0 = 16 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = (16)^2 + 2 (-10) h \Rightarrow 0 = 256 - 20 h \Rightarrow h = \frac{256}{20} = 12.8 \, \text{m} \).

The maximum height is \( 12.8 \, \text{m} \).

12.8 m
16 m
8 m
10 m
1

A runner moves at a constant speed of \( 4 \, \text{m/s} \) for \( 75 \, \text{s} \). What is the distance covered?

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

Here, \( v = 4 \, \text{m/s} \), \( t = 75 \, \text{s} \).

Substitute: \( x = 4 \cdot 75 = 300 \, \text{m} \).

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

250 m
275 m
325 m
300 m
4

A truck traveling at \( 36 \, \text{km/h} \) stops in \( 5 \, \text{s} \) with uniform deceleration. What is the distance covered before stopping?

Convert speed: \( 36 \, \text{km/h} = 36 \cdot \frac{1000}{3600} = 10 \, \text{m/s} \).

Find \( a = \frac{v - v_0}{t} = \frac{0 - 10}{5} = -2 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 10 \cdot 5 + \frac{1}{2} (-2) (5)^2 = 50 - 25 = 25 \, \text{m} \).

The distance covered is \( 25 \, \text{m} \).

15 m
20 m
25 m
30 m
3

A ball is dropped from a height of \( 20 \, \text{m} \). What is its velocity just before hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Use \( v^2 = v_0^2 + 2 g y \). Here, \( v_0 = 0 \), \( g = 10 \, \text{m/s}^2 \), \( y = 20 \, \text{m} \).

Substitute: \( v^2 = 0 + 2 \cdot 10 \cdot 20 = 400 \Rightarrow v = \sqrt{400} = 20 \, \text{m/s} \).

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

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

A car decelerates uniformly from \( 45 \, \text{m/s} \) to rest over a distance of \( 112.5 \, \text{m} \). What is the deceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v = 0 \), \( v_0 = 45 \, \text{m/s} \), \( x = 112.5 \, \text{m} \).

Substitute: \( 0 = (45)^2 + 2 a (112.5) \Rightarrow 0 = 2026 + 225 a \Rightarrow a = -9 \, \text{m/s}^2 \).

Magnitude of deceleration is \( 9 \, \text{m/s}^2 \).

6 m/s²
9 m/s²
4.5 m/s²
12 m/s²
2

A ball is thrown upwards at \( 25 \, \text{m/s} \) from a \( 60 \, \text{m} \) cliff. What is its speed when it hits the ground? (Take \( g = 10 \, \text{m/s}^2 \))

\( v^2 = v_0^2 + 2 g h \), \( v^2 = (25)^2 + 2 \cdot 10 \cdot 60 = 625 + 1200 = 1825 \).

\( v = \sqrt{1825} \approx 42.72 \, \text{m/s} \).

40 m/s
45 m/s
42 m/s
42.72 m/s
4

A truck travels at a constant speed of \( 60 \, \text{km/h} \) for \( 15 \, \text{minutes} \). What is the distance covered?

Convert speed: \( 60 \, \text{km/h} = 60 \cdot \frac{1000}{3600} = \frac{50}{3} \, \text{m/s} \).

Time = \( 15 \, \text{min} = 15 \cdot 60 = 900 \, \text{s} \).

Distance \( x = v t = \frac{50}{3} \cdot 900 = 15000 \, \text{m} = 15 \, \text{km} \).

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

15 km
10 km
20 km
12 km
1

A car decelerates uniformly from \( 18 \, \text{m/s} \) to rest in \( 6 \, \text{s} \). What is the distance covered?

Use \( a = \frac{v - v_0}{t} = \frac{0 - 18}{6} = -3 \, \text{m/s}^2 \).

Distance \( x = v_0 t + \frac{1}{2} a t^2 = 18 \cdot 6 + \frac{1}{2} (-3) (6)^2 = 108 - 54 = 54 \, \text{m} \).

The distance covered is \( 54 \, \text{m} \).

36 m
72 m
18 m
54 m
4

A car moves at \( 20 \, \text{m/s} \) and decelerates at \( 2 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then accelerates at \( 1 \, \text{m/s}^2 \) until it reaches \( 20 \, \text{m/s} \) again. What is the total distance covered?

Phase 1: \( v = 20 - 2 \cdot 5 = 10 \, \text{m/s} \), \( x_1 = 20 \cdot 5 - \frac{1}{2} \cdot 2 \cdot (5)^2 = 100 - 25 = 75 \, \text{m} \).

Phase 2: \( 20 = 10 + 1 \cdot t \Rightarrow t = 10 \, \text{s} \), \( x_2 = 10 \cdot 10 + \frac{1}{2} \cdot 1 \cdot (10)^2 = 100 + 50 = 150 \, \text{m} \).

Total distance = \( 75 + 150 = 225 \, \text{m} \).

200 m
250 m
210 m
225 m
4

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