Question 1.11: A car moves between two sets of traffic lights, stopping at ...
A car moves between two sets of traffic lights, stopping at both. Its speed v ms^{-1} at time t s is modelled by
v = \frac{1}{20} t (40 – t), 0 ≤ t ≤ 40 .
Find the times at which the car is stationary and the distance between the two sets of traffic lights.
The car is stationary when v = 0. Substituting this into the expression for the speed gives
0 = \frac{1}{20} t (40 – t)
t = 0 or t = 40 .
These are the times when the car starts to move away from the first set of traffic lights and stops at the second set.
The distance between the two sets of lights is given by
Distance = \int_{0}^{40}{\frac{1}{20} t (40 – t) dt}
Distance = \frac{1}{20} \int_{0}^{40}{ 40t – t^{2} dt}
Distance = \frac{1}{20} \left[ 20t^{2} – \frac{t^{3}}{3}\right]^{40}_{0}
= 533.3 m
