**Answer:**

There will be a collision 11.4 s after Sue observes the van and hit the brakes.

The collision will occur at 212 m inside the tunnel.

**Explanation:**

The position and velocity of the Sue´s car can be calculated as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time at which Sue´s car and the van meet. At that time, their position will be the same. Let´s place the origin of the system of reference at the point where Sue starts decelerating so that the initial position for Sue is 0 m. Then:

Position of Sue´s car = position of the van

Sue´s car position:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 30.0 m/s · t - 1/2 · 2.00 m/s² · t²

Position of the van:

Since the van travels at constant velocity, a = 0. Then:

x = x0 + v · t

x = 155 m + 5.00 m/s · t

Position of Sue´s car = position of the van

30.0 m/s · t - 1/2 · 2 m/s² · t² = 155 m + 5.00 m/s · t

Let´s solve the equation for t:

30.0 m/s · t - 1.00 m/s² · t² = 155 m + 5.00 m/s · t

- 1.00 m/s² · t² + 30 m/s · t - 5.00 m/s · t - 155 m = 0

-1.00 m/s² · t² + 25 m/s · t - 155 m = 0

Solving the quadratic equation

t = 11.4 s and t = 13.6 s

Let´s consider the lower value because if the collision occurs at t = 11.4, it will not occur at a later time again.

Let´s calculate the velocity of Sue´s car at t = 11.4. If it is greater than zero, then there will be a collision.

v = v0 + a · t

v = 30.0 m/s - 2.00 m/s² · 11.4 s

v = 7.2 m/s

Then, there will be a collision 11.4 s after Sue observes the van and hit the brakes.

The distance inside the tunnel at which the collision occurs can be calculated using the equation of position of the van (or Sue´s car) at t = 11.4 s.

The position of the van at t = 11.4 s is calcualted as follows:

x = x0 + v · t

x = 155 m + 5.00 m/s · 11.4 s

x = 212 m

The collision will occur at 212 m inside the tunnel.