\includegraphics{figure_2}
At time \(t = 0\) a small package is projected from a point \(B\) which is 2.4 m above a point \(A\) on horizontal ground. The package is projected with speed 23.75 m s\(^{-1}\) at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{4}{3}\). The package strikes the ground at the point \(C\), as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.
- Find the time taken for the package to reach \(C\).
[5]
A lorry moves along the line \(AC\), approaching \(A\) with constant speed 18 m s\(^{-1}\). At time \(t = 0\) the rear of the lorry passes \(A\) and the lorry starts to slow down. It comes to rest \(T\) seconds later. The acceleration, \(a\) m s\(^{-2}\) of the lorry at time \(t\) seconds is given by
$$a = -\frac{1}{4}t^2, \quad 0 \leq t \leq T.$$
- Find the speed of the lorry at time \(t\) seconds.
[3]
- Hence show that \(T = 6\).
[3]
- Show that when the package reaches \(C\) it is just under 10 m behind the rear of the moving lorry.
[5]
END