Speedy Sue, driving at 32.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.50 m/s. Sue applies her brakes but can accelerate only at −1.60 m/s2 because the road is wet. Will there be a collision

Sue decelerates at a constant rate, so we can use kinematic equations, also known as SUVAT equations, to model her motion. In this case, we will use two equations:

v = u + at

v² = u² + 2as

where

s is the displacement (distance and direction)
u is the initial velocity (speed and direction)
v is the final velocity (speed and direction)
a is the acceleration
t is the time

In this case, u = 32.0 m/s, v = 5.50 m/s, and a = -1.60 m/s². Use the equations to find the time it takes and the distance traveled for Sue to match the van's speed.

v = u + at

5.50 = 32.0 + (-1.60) t

t = 16.56 s

v² = u² + 2as

(5.50)² = (32.0)² + 2 (-1.60) s

s = 311 m

Now check how far the van travels during that time.

(5.50 m/s) (16.56 s) = 91.09 m

The van was originally 160 m in front of Sue, so by the time she slows down, the van's position will be 160 + 91 = 251 meters from Sue's starting point.

Sue's position will be at 311 meters, and the van's position will be at 251 meters. Therefore, Sue will hit the van before reaching the van's speed.