The safe stopping distance, d, in metres, of a Harold Dobson motorcycle on wet pavement is given by the function d(s)= 0.02(3s^2 + 20s), where s is the speed of the motorcycle in metres per second. Find the speed at which the safe stopping distance is 50 metres.

Answer 1

set 0.02(3s^2 + 20s) = 50
3s^2 + 20s = 2500
3s^2 + 20s - 2500 = 0

Use the quadratic formula, reject the negative answer.

Answer 2

0.02(3s^2+20s)=50
3s^2+20s=2500
3s^2+20s-2500=0
x=25.725992956937827
=25