(a) Given that m = i – i, n = 2i + 3j and 2m + n – r = 0, find |r|
(b) The distance, S metres of a moving particle at any time tseconds is given by
S = 3t – (frac{t^3}{3}) + 9
Find the;
(i) time
(ii) distance travelled
When the particle is momentarily at rest
Explanation
(a) Substituting for m and n to get 2(i – j) + 2i + 3j – r = 0 and solving for r to get r = 4i + j
Thus |r| = (sqrt{4^2 + 1^2}) = (sqrt{17}) = 4.1231
(b)(i) Given that S = 3t – (frac{r^3}{3}) + 9, they took the derivative of S with respect to t and equated to zero to obtain (frac{d_s}{d_t}) = 3 – t(^2) = 0
Solve to get t = (sqrt{3}) = 1.7321 seconds
(b)(ii) Substituting for (t), the distance covered is S = 3(1.7321) – (frac{(1.7321)^3}{3}) + 9
Therefore; 5.1963 – 1.732198389 + 9
= 12.4641 metres