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2013 WAEC Further Mathematics Theory The displacement S metres of a particle from a fixed point O at time t...

Further Mathematics
WAEC 2013

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 \(0 \leq x \leq 5\).

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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. 

 


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