Question

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshine. When he recovers, he sees that he is fast overtaking a camper ahead in his lane moving at the slower speed of 15.1 m/s. He hits the brakes as fast as he can (his reaction time is 0.31 s). If he can decelerate at 2.5 m/s2, what is the minimum distance between the driver and the camper when he first sees it so that they do not collide

Answer 1

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

$v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s$

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

$x=vt=(15.1m/s)(13.49s)=203.69m$   (1)

Next you calculate the distance that teacher has traveled for 13.6s:

$x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m$  (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

$x_{min}=692.88m-203.69m=489.19m$