Class IX Science – Motion (Practice Flashcards)

18 question-and-answer flashcards covering key numerical results, formulas, and conceptual ideas from the Motion assignment.

What distance does a particle cover when it completes three revolutions in a circle of diameter 5 m?

15 π m (≈ 47.1 m).

What is the displacement of the particle after the same three complete revolutions?

0 m, because it returns to the starting point.

If a body is thrown vertically upward to a maximum height h and comes back to the ground, what total distance does it travel?

2 h (upward h + downward h).

What is the body’s displacement after returning to the launch point?

0 m.

An object travels 16 m in 4 s and the next 16 m in 2 s. What is its average speed?

5.33 m s⁻¹ (total 32 m in 6 s).

Vishnu swims 180 m (there and back) in a 90 m pool in 60 s. What is his average speed?

3 m s⁻¹.

What is Vishnu’s average velocity for the same swim?

0 m s⁻¹, because his displacement is zero (he ends where he started).

How far will a train moving at 120 km h⁻¹ travel in 30 s?

1000 m (velocity 33.3 m s⁻¹ × 30 s).

A body rolls with 0.5 m s⁻¹ and decelerates at 0.05 m s⁻². How long until it stops?

10 s.

A car accelerates from 36 km h⁻¹ to 70 km h⁻¹ in 5 s. What is its acceleration?

≈ 1.9 m s⁻².

If the same car (70 km h⁻¹) stops uniformly in 20 s, what is its retardation?

≈ 0.97 m s⁻² (opposite in sign to acceleration).

How do you obtain speed from a position–time graph?

By finding the slope (change in position ÷ change in time) of the graph’s segment.

How do you find acceleration from a velocity–time graph?

By taking the slope (change in velocity ÷ change in time) of the v–t graph.

A body starts from rest with uniform acceleration 10 m s⁻². How much distance does it cover in 2 s?

20 m.

A walker goes 6 km east in 1 h, then returns the same 6 km at 4 km h⁻¹. What is his average speed?

4.8 km h⁻¹ (12 km in 2.5 h).

What is the walker’s average velocity for the whole trip?

0 km h⁻¹ (net displacement is zero).

A train starts from rest and reaches 72 km h⁻¹ in 5 min. What is its acceleration?

0.067 m s⁻² (20 m s⁻¹ in 300 s).

How far does the train travel during this 5-minute acceleration?

≈ 3.0 km (about 2997 m).