2 A car of mass 1200 kg travels along a road for two minutes during which time it rises a vertical distance of 60 m and does \(1.8 \times 10 ^ { 6 } \mathrm {~J}\) of work against the resistance to its motion. The speeds of the car at the start and at the end of the two minutes are the same.

1. Calculate the average power developed over the two minutes. The car now travels along a straight level road at a steady speed of \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) while developing constant power of 13.5 kW .
2. Calculate the resistance to the motion of the car. How much work is done against the resistance when the car travels 200 m ? While travelling at \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the car starts to go down a slope inclined at \(5 ^ { \circ }\) to the horizontal with the power removed and its brakes applied. The total resistance to its motion is now 1500 N .
3. Use an energy method to determine how far down the slope the car travels before its speed is halved. Suppose the car is travelling along a straight level road and developing power \(P \mathrm {~W}\) while travelling at \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) with acceleration \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\) against a resistance of \(R \mathrm {~N}\).
4. Show that \(P = ( R + 1200 a ) v\) and deduce that if \(P\) and \(R\) are constant then if \(a\) is not zero it cannot be constant.