Speedy Sue, driving at 35.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 175 m ahead traveling at 5.50 m/s. Sue applies her brakes but can accelerate only at −1.40 m/s2 because the road is wet. Will there be a collision?

Question

Yes
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. (If no, enter "0" for the time.)
distance - ???
time - ????

Answer 1

using Calculus:
a = -1.4 m/s^2
v = -1.4t + c
when t = 0, v = 35, so c = 35

v = -1.4t + 35

when she stops (with nothing in front of her), v = 0
t = 35/1.4 = 25 seconds
so it would take her 25 seconds to stop

s = -.7t^2 + 35t + k
when t = 0, s = 0, so k = 0

s = -.7t^2 + 35t
so in the 25 seconds needed to stop, she would have gone
-.7(25^2) + 35(25) or 437.5 m

in the meantime, the van would have covered an additional 25(5.5)
or 137.5 m, plus the original distance of 175 ahead would put the van 312.5 m ahead of the place when she first applied the brakes.
So she will rear end the van!

when does this happen?
when -.7t^2 + 35t = 175+5.5t
.7t^2 - 29.5t + 175 = 0
t = 50/7 seconds or 35 seconds, rejecting the 35 s,

clearly she will rear end at 50/7 s or appr 7.14 seconds

distance travelled by both = 175 + 5.5(50/7) = appr 214.3 m

Answer 2

yes