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WAEC 2022 · Paper 2 · Q14

  1. (a)

    A particle initially at rest moves in a straight line with an acceleration of (10t−4t2) m s−2(10t - 4t^2)\text{ m s}^{-2}. Find the: (i) velocity of the particle after tt seconds; (ii) average acceleration of the particle during the 4th second.

  2. (b)

    A load of mass 120 kg120\text{ kg} is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards: (i) at a constant velocity; (ii) with an acceleration of 3 m s−23\text{ m s}^{-2}. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. v=∫(10t−4t2) dtv = \displaystyle\int (10t - 4t^2)\,dt
    =5t2−43t3+k= 5t^2 - \frac43t^3 + k.
  2. It starts at rest, so k=0k = 0 and v=5t2−43t3v = 5t^2 - \frac43t^3.

(ii)

  1. The 4th second runs from t=3t = 3 to t=4t = 4.
  2. v(3)=45−36=9v(3) = 45 - 36 = 9 and v(4)=80−2563=−163v(4) = 80 - \frac{256}{3} = -\frac{16}{3}.
  3. Average acceleration =v(4)−v(3)1= \dfrac{v(4) - v(3)}{1}
    =−163−9= -\dfrac{16}{3} - 9
    =−433= -\dfrac{43}{3}
    =−1413 m s−2= -14\frac13\text{ m s}^{-2}.

(b)(i)

  1. At constant velocity there is no acceleration: R−120×10=0R - 120 \times 10 = 0, so R=1200R = 1200 N.

(ii)

  1. Accelerating upwards: R−1200=120×3R - 1200 = 120 \times 3, so R=1560R = 1560 N.

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