A cyclist and her bicycle have a combined mass of 100 kg . She is working at a constant rate of 80 W and is moving in a straight line on a horizontal road. The resistance to motion is proportional to the square of her speed. Her initial speed is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and her maximum possible speed under these conditions is \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When she is at a distance \(x \mathrm {~m}\) from a fixed point \(O\) on the road, she is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) away from \(O\).
Show that
$$v \frac { \mathrm {~d} v } { \mathrm {~d} x } = \frac { 8000 - v ^ { 3 } } { 10000 v }$$
Find the distance she travels as her speed increases from \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Use the trapezium rule, with 2 intervals, to estimate how long it takes for her speed to increase from \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).