WAEC 2016 · Paper 2 · Q15

  1. (a)

    Given that m=6i+8j\mathbf m = 6\mathbf i + 8\mathbf j and n=−8i+73j\mathbf n = -8\mathbf i + \frac73\mathbf j, find, correct to two decimal places, the magnitudes and directions (bearings) of m\mathbf m and n\mathbf n.

    Show the answer

    ∣m∣=10.00|\mathbf m| = 10.00 on 036.87∘036.87^\circ; ∣n∣=8.33|\mathbf n| = 8.33 on 286.26∘286.26^\circ

  2. (b)

    A car travelling at 30 m s−130\text{ m s}^{-1} is brought to rest in a distance of 50 m50\text{ m}. Calculate: (i) its acceleration; (ii) the time taken for it to come to rest.

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

Worked solution (try it first)

(a)

  1. ∣m∣=36+64=10.00|\mathbf m| = \sqrt{36 + 64} = 10.00.
  2. m\mathbf m points 6 east and 8 north: tan⁡−168=36.87∘\tan^{-1}\frac68 = 36.87^\circ from north, a bearing of 036.87∘036.87^\circ.
  3. ∣n∣=64+499|\mathbf n| = \sqrt{64 + \frac{49}{9}}
    =6259= \sqrt{\frac{625}{9}}
    =253= \frac{25}{3}
    ≈8.33\approx 8.33.
  4. n\mathbf n points 8 west and 73\frac73 north: tan⁡−187/3=73.74∘\tan^{-1}\frac{8}{7/3} = 73.74^\circ west of north.
  5. So the bearing is 360∘−73.74∘=286.26∘360^\circ - 73.74^\circ = 286.26^\circ.

(b)(i)

  1. v2=u2+2asv^2 = u^2 + 2as: 0=900+100a0 = 900 + 100a, so a=−9 m s−2a = -9\text{ m s}^{-2}.

(ii)

  1. v=u+atv = u + at: 0=30−9t0 = 30 - 9t, so t=309≈3.33 st = \frac{30}{9} \approx 3.33\text{ s}.

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