Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 175 m ahead traveling at 5.40 m/s. Sue applies her brakes but can accelerate only at −1.80 m/s2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs.

Question

mathhelper · Answer

for Sue:
a = -1.8
v = -1.8t + c
when t = 0 , (she applied the brake), her v = 34 m/s
34 = -1.8(0) + c
c = 35.8

Sue's velocity equation: v = -1.8t + 35.8
Sue's distance equation:
d = -.9t^2 + 35.8t + k
let's look at the distance when she applied the break, that is, when t=0
d = 0 + 0 + k, so k = 0 and

d = -.9t^2 + 35.8t

after t seconds, the car in front of her will be (175 + 5.4t)
after t seconds Sue will have travelled -.9t^2 + 35.8t m
If there is a collision those must be equal, that is
-.9t^2 + 35.8t = 175 + 5.4t
-.9t^2 + 30.4t - 175 = 0
.9t^2 - 30.4t + 175 = 0
t = (30.4 ± √294.16)/1.8
= appr 7.138 or 26.4 , we will reject 26.4 since Sue has already collided

when t = 7.138
d = 175 + 5.4(7.138) = 213.5 m into the tunnel

check my arithmetic