The displacement S metres of a particle from a fixed point O at time t seconds is given by (S = t^{2} – 6t + 5).
(a) On a graph sheet, draw a displacement- time graph for the interval (0 leq x leq 6).
(b) From the graph, find the : (i) time at which the velocity is zero ; (ii) average velocity over the interval (0 leq x leq 4) ; (iii) total distance covered in the interval (0 leq x leq 5).
Explanation
(a) (s = t^{2} – 6t + 5)
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| s (m) | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
Scale : On time- axis, 2 cm rep 2 s
On displacement, 2 cm rep 2m.
(b)(i) Velocity = (frac{mathrm d s}{mathrm d t} = Gradient)
Velocity = 0 m/s when t = 3s
(ii) Average velocity = (frac{text{total costume}}{text{total time}})
(frac{5 + 4 + (4 – 3)}{4})
= (frac{10}{4} = 2.5 m/s).
(iii) Total distance covered in the interval (0 leq x leq 5)
= 5 + 4 + 4 = 13m.