(a) P(-1, 4), Q(2, 3), R(x, y) and S(-2, 3) are the verticles of a parallelogram. Find the value of x and y. (b) A particle starts from rest and moves in a straight line. It attains a velocity of 20ms\(^{-1}\) after travelling a distance of 8 metres. Calculate; (ii) Iis acceleration (ii) the time taken to travel 40 metres
Explanation
\(\over{PS}\) = \(\over{OS}\) - \(\over{OP}\) = (\(^{-2}_3\)) - (\(^{-1}_4\)) = (\(^{-1}_{-1}\)) Similarly \(\over{QR}\) = \(\over{OR}\) - \(\over{OQ}\) = (\(^x_y\)) - (\(^2_3\)) = (\(^{x - 2}_{y - 3}\)) Since \(\over{PS}\) = \(\over{QR}\), it flows that (\(^{x - 2}_{y - 3}\)) = (\(^{-1}_{-1}\)). Therefore, x - 2 = -1 which solved will yield x = 1. Also, y - 3 = -1 which implies that y = 2
(b)(i), Since the particle starts from rest, the initial velocity (u) = 0 and substituting for final velocity (v) and distance (s) into the formula v\(^2\) = u\(^2\) + 2as to obtain 20\(^2\) = 0 + 2a(8) and when simplifies and solved for acceleration (a), to get a = 25ms\(^{-2}\) (b)(ii), S = ut + \(\frac{1}{2}\)at\(^2\), where s = 40, u = 0 and a = 25 Substituting these values into the equation we obtain 40 = 0 + \(\frac{1}{2}\) x 25t\(^2\) and solving for t gave t = 1.79 seconds.
