17 A cyclist accelerates from rest for 5 seconds then brakes for 5 seconds, coming to rest at the end of the 10 seconds. The total mass of the cycle and rider is \(m \mathrm {~kg}\), and at time \(t\) seconds, for \(0 \leqslant t \leqslant 10\), the cyclist's velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
A resistance to motion, modelled by a force of magnitude 0.1 mvN , acts on the cyclist during the whole 10 seconds.
- Explain why modelling the resistance to motion in this way is likely to be more realistic than assuming this force is constant.
During the braking phase of the motion, for \(5 \leqslant t \leqslant 10\), the brakes apply an additional constant resistance force of magnitude \(2 m \mathrm {~N}\) and the cyclist does not provide any driving force.
- Show that, for \(5 \leqslant t \leqslant 10 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = - 2\).
- Solve the differential equation in part (b).
- Hence find the velocity of the cyclist when \(t = 5\).
During the acceleration phase ( \(0 \leqslant t \leqslant 5\) ), the cyclist applies a driving force of magnitude directly proportional to \(t\).
- Show that, for \(0 \leqslant t \leqslant 5 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = \lambda t\), where \(\lambda\) is a positive constant.
- Show by integration that, for \(0 \leqslant t \leqslant 5 , v = 10 \lambda \left( t - 10 + 10 \mathrm { e } ^ { - 0.1 t } \right)\).
- Hence find \(\lambda\).
- Find the total distance, to the nearest metre, travelled by the cyclist during the motion.