WAEC 2019 · Paper 2 · Q15

  1. (a)

    Given that m=i−j\mathbf m = \mathbf i - \mathbf j, n=2i+3j\mathbf n = 2\mathbf i + 3\mathbf j and 2m+n−r=02\mathbf m + \mathbf n - \mathbf r = \mathbf 0, find ∣r∣|\mathbf r|.

  2. (b)

    The distance, SS metres, of a moving particle at any time tt seconds is given by S=3t−t33+9S = 3t - \dfrac{t^3}{3} + 9. Find the: (i) time; (ii) distance travelled, when the particle is momentarily at rest.

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

Worked solution (try it first)

(a)

  1. r=2m+n\mathbf r = 2\mathbf m + \mathbf n
    =2(i−j)+(2i+3j)= 2(\mathbf i - \mathbf j) + (2\mathbf i + 3\mathbf j)
    =4i+j= 4\mathbf i + \mathbf j.
  2. ∣r∣=42+12|\mathbf r| = \sqrt{4^2 + 1^2}
    =17= \sqrt{17}
    ≈4.123\approx 4.123.

(b)(i)

  1. Momentarily at rest when dSdt=3−t2=0\dfrac{dS}{dt} = 3 - t^2 = 0, so t=3≈1.732t = \sqrt3 \approx 1.732 s (time is positive).

(ii)

  1. S=33−(3)33+9S = 3\sqrt3 - \dfrac{(\sqrt3)^3}{3} + 9
    =33−3+9= 3\sqrt3 - \sqrt3 + 9
    =9+23= 9 + 2\sqrt3
    ≈12.46\approx 12.46 m.

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