(a) A car is moving with a velocity of 10ms\(^{-1}\) It then accelerates at 0.2ms\(^{-2}\) for 100m. Find, correct to two decimal places the time taken by t...
Question 1 Report
(a) A car is moving with a velocity of 10ms\(^{-1}\) It then accelerates at 0.2ms\(^{-2}\) for 100m. Find, correct to two decimal places the time taken by the car to cover the distance.
(b) A particle moves along a straight line such that its distance S metres from a fixed point O is given by S = t\(^2\) - 5t + 6, where t is the time in seconds. Find its:
(i) initial velocity;
(ii) distance when it is momentarily at rest
Answer Details
(a) Given \(u = 10\,\text{ms}^{-1},\ a = 0.2\,\text{ms}^{-2},\ s = 100\,\text{m}\). Use \(s = ut + \tfrac{1}{2}at^2\):
\[100 = 10t + 0.1t^2 \ \Rightarrow\ t^2 + 100t - 1000 = 0\] \[t = \frac{-100 + \sqrt{100^2 + 4000}}{2} = \frac{-100 + \sqrt{14000}}{2} = \frac{-100 + 118.32}{2}\] \[t \approx 9.16\,\text{s}\](b) \(S = t^2 - 5t + 6\), so velocity \(v = \dfrac{dS}{dt} = 2t - 5\).
(i) Initial velocity at \(t = 0\): \(v = 2(0) - 5 = -5\,\text{ms}^{-1}\).
(ii) Momentarily at rest when \(v = 0\): \(2t - 5 = 0 \Rightarrow t = 2.5\,\text{s}\).
\[S = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25\,\text{m}\]The particle is \(0.25\,\text{m}\) from \(O\) (on the negative side).
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