Solve Collision Question: Will Sue's Car Hit the Van?

[physics_geek]

collision question!

Speedy Sue, driving at 32.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling with velocity 5.20 m/s. Sue applies her brakes but can accelerate only at -2.00 m/s2 because the road is wet. Will there be a collision?



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 and enter zero for the time.




i already know there will be a collision
but I am not sure how to calculate how far into the tunnel it will happen

im guessin to use the final position function..but i dunno

 

[CompuChip]

If Sue starts braking immediately after entering the tunnel, how far into the tunnel will she have come to a stop?
How long will it take her to brake? What is then the distance the van has travelled?

 

[baba89]

how to plug in the numbers

I would approach this problem by first analyzing the given information and identifying the key variables involved. In this case, we have Sue's initial speed (32.0 m/s), the van's initial speed (5.20 m/s), and Sue's acceleration (-2.00 m/s2). We also know that the distance between Sue's car and the van is 160 m.

Next, I would use the equations of motion to calculate the time it takes for Sue's car to reach the van's position. We can use the equation s = ut + 1/2at^2, where s is the final position, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get:

160 m = (32.0 m/s)t + 1/2(-2.00 m/s2)t^2

Simplifying, we get the quadratic equation -2.00t^2 + 32.0t - 160 = 0. Solving for t using the quadratic formula, we get t ≈ 5.57 seconds.

This means that it will take Sue's car approximately 5.57 seconds to reach the van's position. Now, we can use the equation v = u + at to calculate the final velocity of Sue's car at the time of the collision. Plugging in the values, we get:

v = 32.0 m/s + (-2.00 m/s2)(5.57 s) = 21.86 m/s.

Since the van is traveling at a speed of 5.20 m/s, the relative velocity between Sue's car and the van at the time of collision is 21.86 m/s - 5.20 m/s = 16.66 m/s.

To determine the distance into the tunnel where the collision occurs, we can use the equation s = ut + 1/2at^2 again. However, this time we are solving for s. Plugging in the values, we get:

s = (32.0 m/s)(5.57 s) + 1/2(-2.00 m/s2)(5.57 s)^2 = 88.96 m.

Therefore, the collision will occur approximately 88.96 meters into the tunnel. We can also use the equation s = ut + 1/