8 A car of mass 600 kg is driven along a straight horizontal road. The resistance to motion of the car is \(k v ^ { 2 }\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of the car at time \(t\) seconds and \(k\) is a constant.
- When the engine of the car has power 8 kW , show that the equation of motion of the car is
$$600 \frac { \mathrm {~d} v } { \mathrm {~d} t } - \frac { 8000 } { v } + k v ^ { 2 } = 0$$
- When the velocity of the car is \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the engine is turned off.
- Show that the equation of motion of the car now becomes
$$600 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - k v ^ { 2 }$$
- Find, in terms of \(k\), the time taken for the velocity of the car to drop to \(10 \mathrm {~ms} ^ { - 1 }\).