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...
Question 1 Report
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 \(0 \leq x \leq 5\).
Answer Details
(a) Tabulate \( S = t^2 - 6t + 5 \) for \( 0 \le t \le 6 \) and plot S against t.
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| S (m) | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
The graph is a parabola (minimum at \(t = 3\)).
(b)(i) Velocity is zero at the turning point of the graph, i.e. \( \dfrac{dS}{dt} = 2t - 6 = 0 \Rightarrow t = 3\text{ s}. \)
(ii) Average velocity over \( 0 \le t \le 4 \): \[ \frac{S(4) - S(0)}{4 - 0} = \frac{-3 - 5}{4} = -2\text{ ms}^{-1}. \]
(iii) Total distance over \( 0 \le t \le 5 \): the particle moves from \(S=5\) down to the minimum \(S(3) = -4\) (distance \(9\text{ m}\)), then back up to \(S(5) = 0\) (distance \(4\text{ m}\)):
\[ \text{Total distance} = 9 + 4 = 13\text{ m}. \]
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