One-Dimensional Motion Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A cyclist moves at a constant speed of \( 7 \, \text{m/s} \) for \( 60 \, \text{s} \). What is the distance covered?

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

Here, \( v = 7 \, \text{m/s} \), \( t = 60 \, \text{s} \).

Substitute: \( x = 7 \cdot 60 = 420 \, \text{m} \).

The distance covered is \( 420 \, \text{m} \).

420 m
350 m
500 m
400 m
1

A car moving at \( 30 \, \text{m/s} \) decelerates uniformly to rest in \( 6 \, \text{s} \). What is the distance covered?

Find \( a = \frac{v - v_0}{t} = \frac{0 - 30}{6} = -5 \, \text{m/s}^2 \).

Use \( x = v_0 t + \frac{1}{2} a t^2 = 30 \cdot 6 + \frac{1}{2} (-5) (6)^2 = 180 - 90 = 90 \, \text{m} \).

The distance covered is \( 90 \, \text{m} \).

60 m
120 m
90 m
150 m
3

A stone falls from a height \( h \) and covers \( 78.4 \, \text{m} \) in the last \( 2 \, \text{s} \) of its fall. What is the total time of fall? (Take \( g = 9.8 \, \text{m/s}^2 \))

Total time = \( t \), last 2 s: \( h = \frac{1}{2} \cdot 9.8 \cdot t^2 \), \( h - 78.4 = \frac{1}{2} \cdot 9.8 \cdot (t - 2)^2 \).

Distance in last 2 s: \( 78.4 = 4.9 [t^2 - (t - 2)^2] \Rightarrow 16 = t^2 - (t^2 - 4t + 4) \Rightarrow 16 = 4t - 4 \Rightarrow t = 5 \, \text{s} \).

4 s
6 s
5 s
4.5 s
3

A ball is thrown upwards with a speed of \( 42 \, \text{m/s} \). What is the time to reach the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

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

Here, \( v_0 = 42 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = 42 - 10 t \Rightarrow 10 t = 42 \Rightarrow t = 4.2 \, \text{s} \).

The time taken is \( 4.2 \, \text{s} \).

3.8 s
4 s
4.5 s
4.2 s
4

A ball is thrown upwards with a speed of \( 60 \, \text{m/s} \). What is the time to reach the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

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

Here, \( v_0 = 60 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = 60 - 10 t \Rightarrow 10 t = 60 \Rightarrow t = 6 \, \text{s} \).

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

5 s
7 s
8 s
6 s
4

A ball is thrown upwards at \( 15 \, \text{m/s} \) from a \( 45 \, \text{m} \) tower. What is its speed when it passes \( 15 \, \text{m} \) below the tower’s top? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement \( y = -15 \, \text{m} \), \( v^2 = (15)^2 + 2 \cdot 10 \cdot 15 = 225 + 300 = 525 \).

\( v = \sqrt{525} \approx 22.91 \, \text{m/s} \).

20 m/s
25 m/s
22 m/s
22.91 m/s
4

A ball is thrown upwards with a speed of \( 52 \, \text{m/s} \). What is the time to reach the maximum height? (Take \( g = 10 \, \text{m/s}^2 \))

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

Here, \( v_0 = 52 \, \text{m/s} \), \( a = -10 \, \text{m/s}^2 \).

Substitute: \( 0 = 52 - 10 t \Rightarrow 10 t = 52 \Rightarrow t = 5.2 \, \text{s} \).

The time taken is \( 5.2 \, \text{s} \).

5 s
4.8 s
5.5 s
5.2 s
4

A car moving at \( 72 \, \text{km/h} \) decelerates uniformly and stops in \( 8 \, \text{s} \). What is the magnitude of deceleration?

Convert speed: \( 72 \, \text{km/h} = 72 \cdot \frac{1000}{3600} = 20 \, \text{m/s} \).

Use \( a = \frac{v - v_0}{t} \). Here, \( v = 0 \), \( v_0 = 20 \, \text{m/s} \), \( t = 8 \, \text{s} \).

Substitute: \( a = \frac{0 - 20}{8} = -2.5 \, \text{m/s}^2 \).

Magnitude of deceleration is \( 2.5 \, \text{m/s}^2 \).

2.5 m/s²
3 m/s²
2 m/s²
4 m/s²
1

A cyclist moves at a constant speed of \( 9 \, \text{km/h} \) for \( 20 \, \text{minutes} \). What is the distance covered?

Convert speed: \( 9 \, \text{km/h} = 9 \cdot \frac{1000}{3600} = 2.5 \, \text{m/s} \).

Time = \( 20 \, \text{min} = 20 \cdot 60 = 1200 \, \text{s} \).

Distance \( x = v t = 2.5 \cdot 1200 = 3000 \, \text{m} = 3 \, \text{km} \).

The distance covered is \( 3 \, \text{km} \).

2 km
2.5 km
3.5 km
3 km
4

A truck moves at a constant speed of \( 25 \, \text{m/s} \) for \( 20 \, \text{s} \). What is the distance covered?

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

Here, \( v = 25 \, \text{m/s} \), \( t = 20 \, \text{s} \).

Substitute: \( x = 25 \cdot 20 = 500 \, \text{m} \).

The distance covered is \( 500 \, \text{m} \).

500 m
400 m
600 m
450 m
1

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0