(a) A car is moving with a velocity of 10ms\(^{-1}\) It then accelerates at 0.2ms\(^{-2}\)...
(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
Explanation
(a) u = 10ms\(^{-1}\), a = 0.2m\(^{-2}\), s = 100m, t = ?
using S = ut + \(\frac{1}{2} at^2\)
100 = 10t + \(\frac{1}{2}(0.2)t^2\)
100 = 10t + 0.1t\(^2\)
0.1t\(^2\) + 10t - 100 = 0
t = \(\frac{-10 \pm \sqrt{10^2 - 4(0.1)(-100)}}{2(0.1)}\)
t = \(\frac{-10\pm \sqrt{100 + 40}}{0.2}\)
t = \(\frac{-10 \pm \sqrt{140}}{0.2}\)
t = \(\frac{-10 \pm 11.83}{0.2}\)
t = \(\frac{-10 - 11.82}{2}\)
t = \(\frac{1.83}{0.2}\)
t = \(\frac{21.83}{2}\)
t = 9.15 seconds
(b)(i) S = t\(^2\) - 5 + 6
v = \(\frac{ds}{dt}\)
= 2t - 5
at t = 0, y = 2(0) - 5
= - 5ms\(^{-1}\)
The initial velocity is 5 in the negative direction
(ii) When t = 0, S = t\(^2\) - 5t + 6
S = 0\(^2\) - 5(0) + 6
S = 6
The distance is 6m