WAEC 2023 · Paper 1 · Q19

A particle began to move at 27 m s−127\ \text{m s}^{-1} along a straight line with constant retardation of 9 m s−29\ \text{m s}^{-2}. Calculate the time it took the particle to come to a stop.

Worked solution (try it first)
  1. Use v=u+atv = u + at with u=27u = 27, v=0v = 0 and a=−9a = -9 (a retardation is a negative acceleration).
  2. So 0=27−9t0 = 27 - 9t, which gives 9t=279t = 27.
  3. So t=3t = 3 seconds, option B.

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