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NECO 2023 · Paper 1 · Q60

A particle moves in a straight line such that its velocity after tt seconds is (3t+4) m/s(3t + 4)\text{ m/s}. Find the distance travelled in 4 seconds.

Worked solution (try it first)
  1. Distance is the integral of velocity: ∫04(3t+4) dt=[32t2+4t]04\int_0^4 (3t + 4)\,dt = \left[\frac32t^2 + 4t\right]_0^4.
  2. At t=4t = 4: 32×16+16=24+16=40\frac32 \times 16 + 16 = 24 + 16 = 40.
  3. At t=0t = 0: 0.
  4. So the distance is 40 m, option C.

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