Question

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 170 m ahead traveling with velocity 5.10 m/s. Sue applies her brakes but can accelerate only at ?2.00 m/s² because the road is wet. Will there be a collision?

Question 1: Yes or No

If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van, and enter zero for the time.

Question 2: distance ____

                 time______ s

Answer 1

I will assume that the vehicle ahead of Sue continues to move at a constant rate of 5.3 m/s. We will call x=0 the position when Sue applies her brakes.

Sue's equation of motion is:

xs(t)=34m/s t - 1/2 (2m/s/s)t^2

the equation of motion of the other driver is

xd(t) = 170 + 5.3t

you could plot these to see if the curves intersect, or look for a solution by equating the two equations for position

if sue overtakes the driver, xs(t)=xd(t) or

34t-t^2=170+5.3t

collecting terms:

28.7t-t^2-170=0 or

t^2-28.7t+170=0

this is a quadratic with solutions:

t=[28.7+/-sqrt[28.7^2-4(170)]]/2

t=[28.7+/-sqrt[143.7]]/2
t=8.36s and t = 20.34 s

the first time is when sue collides with the driver, the second time would be when the driver (assuming no collision and the driver continued to move at 5.3m/s) would pass the braking sue

so the collision occurs 8.36 s after Sue applies the brakes, in this time Sue travels a distance

x=34t-t^2 = 34(8.36)-(8.36)^2 = 214.3m