Question

7. The following table describes the distance covered by a body moving along a straight line to north direction for the first 10 seconds.
S(m) 0 20 40 60 80 100
t(s) 0 2 4 6 8 10



Based on the information given in the table,
(a) Plot the distance against time graph using square sheet.
(b) Compute the velocity of the car.
(c) Plot the velocity against time graph of the car using square sheet.
(d) Determine the initial position of the car

Answer 1

First read the data as ordered pairs (t, S) in seconds and metres:
(0, 0), (2, 20), (4, 40), (6, 60), (8, 80), (10, 100).

(a) How to plot S vs t on squared paper
- Choose scales: e.g. x-axis (time) 0–10 s: 1 small square = 1 s (10 squares); y-axis (distance) 0–100 m: 1 small square = 10 m (10 squares).
- Label axes: horizontal = t (s), vertical = S (m). Mark units and origin.
- Plot the points above and join them with a straight line. Because the points lie on a straight line the graph will be a straight line through the origin.

(If you need coordinates to plot exactly: t = 0 → S = 0; t = 2 → S = 20; t = 4 → 40; t = 6 → 60; t = 8 → 80; t = 10 → 100.)

(b) Compute the velocity of the car
Velocity = ΔS/Δt. Between any two successive points ΔS = 20 m and Δt = 2 s, so
v = 20/2 = 10 m/s (north).
Using endpoints: v = (100 − 0)/(10 − 0) = 100/10 = 10 m/s.
So the car moves with constant velocity 10 m/s to the north.

(c) How to plot v vs t on squared paper
- Use horizontal axis time 0–10 s (same scale as above). Vertical axis velocity 0–12 m/s (1 small square = 1 m/s for example).
- Plot the velocity value v = 10 m/s at every time and join: you get a horizontal line at v = 10 m/s from t = 0 to t = 10 s.
(You can also list v at each interval: at t = 0–2, v = 10 m/s; 2–4, v = 10 m/s; … all the same.)

(d) Initial position of the car
From the table S(t = 0) = 0 m. So the initial position is 0 m (taken as the origin), i.e. 0 m north of the chosen reference point.