1. A van of mass 900 kg is moving along a straight horizontal road.

The resistance to the motion of the van is modelled as a constant force of magnitude 600 N . The engine of the van is working at a constant rate of 24 kW .
At the instant when the speed of the van is \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the acceleration of the van is \(2 \mathrm {~ms} ^ { - 2 }\)

1. Find the value of \(V\) Later on, the van is towing a trailer of mass 700 kg up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 14 }\)
   The trailer is attached to the van by a towbar, as shown in Figure 3.
   The towbar is parallel to the direction of motion of the van and the trailer. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{52966963-2e62-4361-bcd5-a76322f8621e-20_367_1194_1091_438} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude 600 N . The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 550 N . The towbar is modelled as a light rod.
   The engine of the van is working at a constant rate of 24 kW .
2. Find the acceleration of the van at the instant when the van and the trailer are moving with speed \(8 \mathrm {~ms} ^ { - 1 }\) At the instant when the van and the trailer are moving up the road at \(9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the towbar breaks. The trailer continues to move in a straight line up the road until it comes to instantaneous rest. The distance moved by the trailer as it slows from a speed of \(9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to instantaneous rest is \(d\) metres.
3. Use the work-energy principle to find the value of \(d\).