(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
(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
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Correct Answer: Option
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
(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