Further Mathematics Questions and Answers

$120 kmh^{-1} = \frac{120 \times 1000}{3600} = \frac{100}{3} = 33.3ms^{-1}$

$\frac{\mathrm d s(t)}{\mathrm d t} = v(t)$ and $\frac{\mathrm d v(t)}{\mathrm d t} = a(t)$

$\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}$

$\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)$

$a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}$

Let the power of $2x^{2}$ be t and the power of $\frac{1}{x} \equiv x^{-1}$ = 9 - t.

$(2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}$

Dealing with x alone, we have

$(x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0$

$3t - 9 = 0 \therefore t = 3$

The binomial expansion is then,

$^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \frac{9!}{(9-3)! 3!} \times 2^{3}$

= 84 x 8

= 672

$F_{1} = 7i + 0j$

$F_{2} = (4\cos\theta)i + (4\sin\theta)j$

$9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}$

$9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta$

$81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)$

Recall, $\cos^{2}\theta + \sin^{2}\theta = 1$

$81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta$

$16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857$

$\theta = \cos^{-1} 0.2857  = 73.40°$

$F = mass \times acceleration$ but $accl = \frac{v - u}{t}$

$\therefore F = m(\frac{v - u}{t})$

$Mass = 28g = 0.028kg$

$v = 5.4 ms^{-1}; u = 0; t = 18secs$

$\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N$