One-Dimensional Motion Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A train moving at \( 126 \, \text{km/h} \) decelerates at \( 2 \, \text{m/s}^2 \) for \( 10 \, \text{s} \), then at \( 5 \, \text{m/s}^2 \) until it stops. What is the total distance covered during deceleration?

Speed: \( 126 \, \text{km/h} = 35 \, \text{m/s} \).

Phase 1: \( v = 35 - 2 \cdot 10 = 15 \, \text{m/s} \), \( x_1 = 35 \cdot 10 - \frac{1}{2} \cdot 2 \cdot (10)^2 = 350 - 100 = 250 \, \text{m} \).

Phase 2: \( t = \frac{15}{5} = 3 \, \text{s} \), \( x_2 = 15 \cdot 3 - \frac{1}{2} \cdot 5 \cdot (3)^2 = 45 - 22.5 = 22.5 \, \text{m} \).

Total = \( 250 + 22.5 = 272.5 \, \text{m} \).

272.5 m
270 m
275 m
280 m
1

A stone is thrown upwards at \( 35 \, \text{m/s} \) from a \( 45 \, \text{m} \) cliff. How far below the cliff’s edge is it after \( 8 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Displacement: \( y = 35 \cdot 8 - \frac{1}{2} \cdot 10 \cdot (8)^2 = 280 - 320 = -40 \, \text{m} \).

Distance below cliff = \( 40 + 45 = 85 \, \text{m} \).

85 m
80 m
90 m
75 m
1

A stone falls freely from a height of \( 98 \, \text{m} \). How long does it take to reach the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

Here, \( y = 98 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \).

Substitute: \( 98 = \frac{1}{2} \cdot 9.8 \cdot t^2 \Rightarrow 98 = 4.9 t^2 \Rightarrow t^2 = 20 \Rightarrow t = \sqrt{20} \approx 4.47 \, \text{s} \).

Rounded to one decimal, the time is \( 4.5 \, \text{s} \).

4.5 s
5 s
4 s
6 s
1

Two stones are released from a height of \( 98 \, \text{m} \), \( 1 \, \text{s} \) apart. How long after the second stone is released do they meet? (Take \( g = 9.8 \, \text{m/s}^2 \))

Stone 1: \( y_1 = \frac{1}{2} \cdot 9.8 \cdot (t + 1)^2 \).

Stone 2: \( y_2 = \frac{1}{2} \cdot 9.8 \cdot t^2 \).

Meet when \( y_1 = y_2 \): \( 4.9 (t + 1)^2 = 98 \Rightarrow (t + 1)^2 = 20 \Rightarrow t + 1 = 4.47 \Rightarrow t = 3.47 \, \text{s} \).

Relative: \( y_1 - y_2 = 98 \Rightarrow 4.9 (t^2 + 2t + 1 - t^2) = 98 \Rightarrow 9.8 t + 4.9 = 98 \Rightarrow t = 9.5 \, \text{s} \) (error, use total height).

Correct: They meet when first hits ground, \( t = 4.47 - 1 = 3.47 \, \text{s} \).

3.47 s
3 s
4 s
3.5 s
1

A ball is thrown vertically upwards with a speed of \( 24 \, \text{m/s} \). What is the total time of flight? (Take \( g = 10 \, \text{m/s}^2 \))

Time to max height: \( v = v_0 + a t \), \( 0 = 24 - 10 t \Rightarrow t = 2.4 \, \text{s} \).

Total time = time up + time down = \( 2.4 \, \text{s} + 2.4 \, \text{s} = 4.8 \, \text{s} \).

The total time is \( 4.8 \, \text{s} \).

4.8 s
2.4 s
5 s
4 s
1

A cyclist accelerates from \( 8 \, \text{m/s} \) at \( 2 \, \text{m/s}^2 \) for \( 4 \, \text{s} \), then decelerates at \( 3 \, \text{m/s}^2 \) to \( 10 \, \text{m/s} \). What is the total distance covered?

Phase 1: \( v = 8 + 2 \cdot 4 = 16 \, \text{m/s} \), \( x_1 = 8 \cdot 4 + \frac{1}{2} \cdot 2 \cdot (4)^2 = 32 + 16 = 48 \, \text{m} \).

Phase 2: \( 10 = 16 - 3 t \Rightarrow t = 2 \, \text{s} \), \( x_2 = 16 \cdot 2 - \frac{1}{2} \cdot 3 \cdot (2)^2 = 32 - 6 = 26 \, \text{m} \).

Total = \( 48 + 26 = 74 \, \text{m} \).

70 m
74 m
76 m
72 m
2

A motorcycle accelerates uniformly from rest at \( 2.5 \, \text{m/s}^2 \) for \( 6 \, \text{s} \). What is the final velocity?

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

Here, initial velocity \( v_0 = 0 \), acceleration \( a = 2.5 \, \text{m/s}^2 \), time \( t = 6 \, \text{s} \).

Substitute: \( v = 0 + 2.5 \cdot 6 = 15 \, \text{m/s} \).

The final velocity is \( 15 \, \text{m/s} \).

15 m/s
20 m/s
10 m/s
12 m/s
1

A stone falls freely from a height of \( 78.4 \, \text{m} \). How long does it take to reach the ground? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

Here, \( y = 78.4 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \).

Substitute: \( 78.4 = \frac{1}{2} \cdot 9.8 \cdot t^2 \Rightarrow 78.4 = 4.9 t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4 \, \text{s} \).

The time taken is \( 4 \, \text{s} \).

4 s
2 s
6 s
8 s
1

A body accelerates uniformly from rest to a velocity of \( 20 \, \text{m/s} \) over a distance of \( 50 \, \text{m} \). What is the acceleration?

Use \( v^2 = v_0^2 + 2 a x \). Here, \( v_0 = 0 \), \( v = 20 \, \text{m/s} \), \( x = 50 \, \text{m} \).

Substitute: \( (20)^2 = 0 + 2 a (50) \Rightarrow 400 = 100 a \Rightarrow a = 4 \, \text{m/s}^2 \).

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

2 m/s²
6 m/s²
4 m/s²
8 m/s²
3

A ball is thrown upwards at \( 40 \, \text{m/s} \) from a \( 25 \, \text{m} \) tower. What is the time taken to hit the ground? (Take \( g = 10 \, \text{m/s}^2 \))

\( y = v_0 t - \frac{1}{2} g t^2 \), \( y = -25 \), \( v_0 = 40 \, \text{m/s} \).

\( -25 = 40 t - 5 t^2 \Rightarrow 5 t^2 - 40 t - 25 = 0 \Rightarrow t^2 - 8 t - 5 = 0 \).

Solve: \( t = \frac{8 \pm \sqrt{64 + 20}}{2} = \frac{8 \pm 9.17}{2} \), \( t = 8.585 \, \text{s} \) (positive root).

8 s
9 s
8.585 s
7.5 s
3

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