LIWA

Motion in One Dimension

ICSE Class 9 · Physics · Chapter 2

What this chapter covers

Key points

Questions answered

Distinguish between a scalar and a vector quantity, with two examples of each.
A scalar has magnitude only (mass, speed); a vector has magnitude and direction (displacement, velocity).
A person is sitting in a moving train. Is he at rest or in motion? Explain.
Both, depending on the reference: he is at rest relative to the train and in motion relative to the ground. Rest and motion are relative.
Can displacement be zero when distance is not zero? Give an example.
Yes. A runner completing one lap of a track covers a distance of one lap but has zero displacement, because the start and finish points are the same.
A car covers 30 km at 30 km h-1 and the next 30 km at 60 km h-1. What is its average speed?
Times are 1 h and 0.5 h; average speed = 60 km ÷ 1.5 h = 40 km h-1.
Convert 90 km h-1 into m s-1.
90 × 5/18 = 25 m s-1.
Define acceleration and state its SI unit. What does a negative acceleration mean?
Acceleration is the rate of change of velocity; its SI unit is m s-2. A negative acceleration points opposite to the direction chosen as positive. It is a retardation (the body slows down) when the positive direction is the direction of motion, that is, when a and v have opposite signs; a body moving in the negative direction with a negative acceleration speeds up.
Why can a body moving at constant speed still have variable velocity?
Velocity includes direction; if the direction changes, as on a circular path, the velocity changes even though the speed does not.
What does the slope of a distance–time graph represent? What does a horizontal line on it mean?
The slope gives the speed. A horizontal line means the body is at rest.

Past board exam questions

ICSE 2019 sample paper · 2 marks

If I travel from Mumbai to Pune (150 Km) in 2½ hrs via the Express Highway and return to Mumbai via the old High way (180 km) in 3½ hrs, calculate the average velocity during the entire journey.

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The journey starts and ends at Mumbai, so the total displacement is zero. Average velocity = total displacement ÷ total time = 0 ÷ (2½ + 3½) h = 0 ÷ 6 h = 0. The average velocity is zero. (The average speed would be 330 km ÷ 6 h = 55 km h⁻¹.)

ICSE 2019 sample paper · 2 marks

Define retardation and give an example of a body having this motion.

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Retardation (deceleration) is the rate of decrease of velocity with time, i.e. negative acceleration. Example: a car slowing down when brakes are applied (or a ball thrown vertically upwards while it rises).

ICSE 2019 sample paper · 3 marks

A train is moving at a velocity of 25 ms⁻¹. It is brought to rest by applying the brakes which produces a uniform retardation of 0.5 ms⁻². Calculate (i) the velocity of the train after 10 s (ii) If the mass of the train is 20000 kg then calculate the force required to stop the train.

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u = 25 m s⁻¹, a = −0.5 m s⁻². (i) v = u + at = 25 + (-0.5)(10) = 25 - 5 = 20 m s⁻¹. (ii) F = ma = 20000 × 0.5 = 10000 N (= 1 × 10⁴ N), acting opposite to the motion.

ICSE 2019 sample paper · 4 marks

Using the following velocity time graph of a body answer the following questions. The graph of velocity V (m s⁻¹) against time t (s) is made of straight lines joining these points: | t (s) | 0 | 3 | 5 | 7 | |---|---|---|---|---| | V (m s⁻¹) | 5 | 15 | 15 | 0 | (i) During which time intervals is the body moving with variable velocity? (ii) What is the acceleration of the body during the interval 3 s to 5 s? (iii) What is the displacement of the body in the last four seconds of its motion?

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(i) The velocity changes from 0 s to 3 s (it increases from 5 to 15 m s⁻¹) and from 5 s to 7 s (it decreases from 15 m s⁻¹ to 0). (ii) From 3 s to 5 s the velocity is constant at 15 m s⁻¹, so the acceleration is zero. (iii) The last four seconds are 3 s to 7 s. Displacement = area under the graph = (15 × 2) + ½ × 2 × 15 = 30 + 15 = 45 m.