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...

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(a) Displacement–time graph

The displacement is \(S = t^{2} - 6t + 5\). Compute \(S\) for each whole second in the interval \(0 \le t \le 6\):

t (s)	0	1	2	3	4	5	6
S (m)	5	0	−3	−4	−3	0	5

Plotting these points and joining them with a smooth curve gives the parabola below (scale on the time axis: 2 cm to 1 s; on the displacement axis: 2 cm to 1 m).

(b) Reading the graph

(i) Time at which the velocity is zero.
The velocity is the gradient of the displacement–time graph, \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t}\). The velocity is zero where the tangent to the curve is horizontal, i.e. at the lowest point (turning point) of the parabola. From the graph this occurs at

\[t = 3\ \text{s}.\]

(Check: \(v = \dfrac{\mathrm{d}S}{\mathrm{d}t} = 2t - 6 = 0 \Rightarrow t = 3\ \text{s}.\))

(ii) Average velocity over the interval \(0 \le t \le 4\).
Average velocity is the change in displacement divided by the time taken (the gradient of the chord joining the two end points). From the graph, \(S = 5\) m at \(t = 0\) and \(S = -3\) m at \(t = 4\):

\[\bar v = \frac{S(4) - S(0)}{4 - 0} = \frac{-3 - 5}{4} = \frac{-8}{4} = -2\ \text{m/s}.\]

The average velocity is \(-2\ \text{m/s}\); the negative sign shows the net motion is directed back towards, and past, \(O\). (For interest, the total path length in this interval is \(9 + 1 = 10\) m, so the average speed is \(10/4 = 2.5\) m/s.)

(iii) Total distance covered in the interval \(0 \le t \le 5\).
Distance is the actual length of path travelled, so the two legs of the journey are added as positive lengths. From the graph the particle first moves from \(S = 5\) m down to the turning point \(S = -4\) m at \(t = 3\) s, then rises back to \(S = 0\) m at \(t = 5\) s.

\[\text{Distance down} = |{-4} - 5| = 9\ \text{m}, \qquad \text{Distance up} = |0 - (-4)| = 4\ \text{m}.\]\[\text{Total distance} = 9 + 4 = 13\ \text{m}.\]