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Speedy Sue, driving at 32.0 m/s, enters a one-lane tunnel. She then observes a s

ID: 2167007 • Letter: S

Question

Speedy Sue, driving at 32.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling with velocity 5.40 m/s. Sue applies her brakes but can accelerate only at ?2.00 m/s2 because the road is wet. Will they collide?

Yes?

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 believe they will collide but starting this problem hasn't been easy for me.

Explanation / Answer

Here is how to solve it. The key is to recognize that there is only one time and distance at which they will collide. The second key is to recognize that each driver has their own equation that involves distance and time, and by setting these equal to each other, we can find that intersection point. Use the equation: d = d0 + v0*t + 1/2*a*t^2 t is time, d is position, a is acceleration, d0 is initial position, and v0 is initial velocity Write the equation for both people. For sue: d = 0 + 32t + 1/2*-2t^2 0 because she starts at the beginning of the tunnel For van: d = 155 + 5.1t + 0 0 because he has no acceleration Now we have two equations, each with the variable d and t. If we set these equal to each other, we will come up with the answers that you're looking for. Let's use substitution: Plug what d equals from the second equation (already solve for d) into the first equation. 155 + 5.1t = 32t -1t^2 Solve for t. I'll leave that up to you, it's a quadratic so just use the quadratic equation, but the answer is: t= 8.36 s There are two answers to this quadratic- just pick the first, since there can be no collision at a later time. Then plug that value into the second equation to get the distance: d = 155 + 5.1*8.36 d = 197.63 m So it happens 197.63 m into the tunnel and at 8.36 seconds after Sue enters the tunnel. Hope this helps

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