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WAEC SSCE - Further Mathematics - 2015 (Objective)

Question 1 Report

A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by \(a = (3t - 2) ms^{-2}\). Find the other time when the velocity would be zero.

Answer Details
The acceleration of the particle is given as \(a = (3t - 2) ms^{-2}\). We can integrate this to get the velocity function, \(v(t)\), of the particle: \[\int a dt = \int (3t - 2) dt\] \[v(t) = \frac{3}{2}t^2 - 2t + C\] where C is the constant of integration. Since the particle starts from rest, we have \(v(0) = 0\), which implies that C = 0. Therefore, the velocity function of the particle is given as: \[v(t) = \frac{3}{2}t^2 - 2t\] To find the other time when the velocity would be zero, we need to solve the equation \(v(t) = 0\): \[\frac{3}{2}t^2 - 2t = 0\] \[t(3t - 4) = 0\] This equation has two roots, \(t = 0\) and \(t = \frac{4}{3}\). Since the particle is starting from rest, we discard the root \(t = 0\). Therefore, the other time when the velocity would be zero is \(t = \frac{4}{3}\) seconds. Hence, the answer is \(\frac{4}{3} seconds\).