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WAEC 2011 · Paper 2 · Q8

  1. (a)

    A car moving on a straight road with constant acceleration has a velocity of 20 km h−120\text{ km h}^{-1} at an instant. If 15 minutes later it had a velocity of 50 km h−150\text{ km h}^{-1}, find the acceleration of the car.

  2. (b)

    A particle is projected vertically upwards with a speed of 40 m s−140\text{ m s}^{-1} from a point on the ground. Find the maximum height reached. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

Worked solution (try it first)

(a)

  1. Keep the units in hours: 15 min=14 h15\text{ min} = \frac14\text{ h}.
  2. v=u+atv = u + at: 50=20+14a50 = 20 + \frac14a, so a=120 km h−2a = 120\text{ km h}^{-2}.

(b)

  1. At the top v=0v = 0: 0=402−2(10)s0 = 40^2 - 2(10)s.
  2. 20s=160020s = 1600, so s=80 ms = 80\text{ m}.

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