Speedy sue, driving at 33.0 m/s, enters a one-lane tunnel. she then observes a slow-moving van 160 m ahead traveling at 5.20 m/s. sue applies her brakes but can accelerate only at −1.50 m/s^2 because the road is wet. will there be a collision? 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.)

The collision will occur 7.12 s from the moment Sue sees the van.

The collision will occur 196.9 m inside the tunnel.

Explanation:

The equation for the position of an object moving in a straight line is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where

x = position at time t

x0 = initial position

t = time

v0 = initial velocity

a = acceleration

If the object moves with constant speed, then a = 0 and x = x0 + v · t

If there is a collision, there will be a time "t" at which the position of Sue and the van is the same.

Let´s place the center of the frame of reference at the position of Sue when she sees the van and applies the brakes.

x Sue = x van

x0 + v0 · t + 1/2 · a · t² = x0 + v · t    (x0 Sue = 0, x0 van = 160 m)

33 m/s · t - 1/2 · 1.50 m/s² · t² = 160m + 5.20 m/s · t

33 m/s  · t - 0.75 m/s² · t² = 160m + 5.20 m/s · t

- 0.75 m/s² · t² +  27.8 m/s · t -160 m = 0

t = 7.12 s  ( the other value is t = 29.9, we take the lower one because the collision will occur only once).

The collision will occur 7.12 s from the moment Sue sees the van.

The distance traveled during that time will be:

x = x0 + v0 · t + 1/2 · a · t²

x = 33 m/s · 7.12 s - 1/2 · 1.50 m/s² · (7.12 s)²  

x = 196.9 m

The collision will occur 196.9 m inside the tunnel.