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NECO 2023 · Paper 2 · Q13

A car starts from town AA and accelerates uniformly for 4 minutes until it reaches a speed of 35 ms−135\text{ ms}^{-1}, which it maintains for 30 minutes; it then retards uniformly for 3 minutes to stop at town BB. Calculate the:

  1. (a)

    distance between AA and BB in kilometres;

  2. (b)

    average speed of the car (ms−1\text{ms}^{-1}, 2 d.p.);

  3. (c)

    acceleration and retardation (ms−2\text{ms}^{-2}, 4 d.p.);

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

  4. (d)

    time taken to reach CC, halfway between AA and BB (seconds).

Worked solution (try it first)
  1. Times in seconds: 4 min=240 s4\text{ min} = 240\text{ s}, 30 min=1800 s30\text{ min} = 1800\text{ s} and 3 min=180 s3\text{ min} = 180\text{ s}.

(a)

  1. Distance = area under the velocity–time graph.
  2. 12(240)(35)+1800(35)+12(180)(35)=4200+63 000+3150\frac12(240)(35) + 1800(35) + \frac12(180)(35) = 4200 + 63\,000 + 3150
    =70 350 m= 70\,350\text{ m}, which is 70.35 km70.35\text{ km}.

(b)

  1. Average speed =70 350240+1800+180= \dfrac{70\,350}{240 + 1800 + 180}
    =70 3502220= \dfrac{70\,350}{2220}
    ≈31.69 m s−1\approx 31.69\text{ m s}^{-1}.

(c)

  1. Acceleration =35240= \dfrac{35}{240}
    ≈0.1458 m s−2\approx 0.1458\text{ m s}^{-2}.
  2. Retardation =35180= \dfrac{35}{180}
    ≈0.1944 m s−2\approx 0.1944\text{ m s}^{-2}.

(d)

  1. Halfway is 35 175 m35\,175\text{ m}.
  2. The first 4200 m4200\text{ m} take 240 s240\text{ s}.
  3. The rest, 30 975 m30\,975\text{ m}, is at 35 m s−135\text{ m s}^{-1}: 30 97535=885 s\dfrac{30\,975}{35} = 885\text{ s}.
  4. So CC is reached after 240+885=1125 s240 + 885 = 1125\text{ s} (18 min 45 s).

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