Speedy Sue, driving at 35.0 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 −1.90 m/s2 because the road is wet. Will there be a collision?

Hence, there will be a collision

Explanation:

First we calculate total distance covered by the speedy sue's car before coming to rest:

2as = Vf² - Vi²

where,

a = deceleration = - 1.9 m/s²

s = distance covered = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 35 m/s

Therefore,

2(-1.9 m/s²)s = (0 m/s)² - (35 m/s)²

s = 322.37 m

Now, we calculate time taken by car to stop:

Vf = Vi + at

0 m/s = 35 m/s + (-1.9 m/s²)t

t =  18.42 s

Now, we calculate distance traveled by van in this time:

s₁ = V₁t

where,

s₁ = distance traveled by van = ?

V₁ = speed of van = 5.2 m/s

Therefore,

s₁ = (5.2 m/s)(18.42 s)

s₁ = 95.78 m

Now, for collision to occur, the following relation must be satisfied:

s ≥ 160 m + s₁

using values:

322.37 m > 160 m + 95.78 m

322.37 m > 255.78 m

Hence, there will be a collision