# Homework Help: Tractor Collision problem

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1. Sep 6, 2016

### ChrisBrandsborg

1. The problem statement, all variables and given/known data
Problem 5: Collision course

A huge tractor and a Tesla full of school children come driving along a winding mountain road, in opposite directions. The tractor has a speed of 40.0km/h and the car zooms along with 80.0 km/h. The Tesla suddenly comes around a corner, sees the tractor, and they both immediately start braking, both with constant accelerations of 5.00 m/s2 (opposite to their directions of motion).

a) If the initial distance between the two is 60.0 m, do they hit each other? If so, where, and with what relative speed on impact? If not, what is the distance between the two when they both stop?

In fact, it takes both of them 0.50 s to react to seeing each other, so they only start braking 0.50s after the car comes round the corner.

b) What is the answer to the questions in a) in this case?

As it happens the tractor driver is looking the other way, and doesn’t brake at all, but continues with his original speed.

c) What should the acceleration a of the Tesla be, to have time to stop before being hit by the tractor (still including the 0.5 s delay from question b))?

2. Relevant equations
1. x = (a/2)t^2 + vi*t (+xi)
2. x = (v^2-vi^2)/2a
3. v = vi + at (this might be relevant?)

3. The attempt at a solution

What I have done so far:

I have set the tractor's direction to be positive.
Tractor = A
Tesla = B

vA = 40km/h = 11.1m/s
vB = -80km/h = -22.2m/s
aA = -5m/s^2
aB = 5m/s^2
x = 60m (distance from each other)

Since vA < vB, we know that the tractor will stop before the Tesla will hit it.
So vA-finale = 0m/s

Then I have made a positional function for both cars.

xA = (-vA^2)/2a
xB(t) = (a/2)t^2 + vB*t + x

Then I set xA = xB(t)

(-vA^2)/2a) = (a/2)t^2 + vB*t + x

(-11.1)^2/10 = (5/2)t^2 - 22.2t + 60

=> (5/2)t^2 - 22.2t + 47.7 = 0
Which gives me t = 3.6 or t = 5.2 (but I assume that the only one valid is t = 3.6?) Since that is the time it takes before they hit each other?

Then I insert t = 3.6 into xB(t) to find the position of Tesla (to find out where they "crashed")
x = 12.5m

Does that mean that he drove (60m - 12.5m) = 47.5m before they crashed?
(which means that the tractor drove 12.5m).

It´s here I start getting confused. Which equation do I use now to find the speed on impact?
V = Vi + at? Where I put in 22.2 m/s for Vi and -5m/s^2 for a, and 3.6 for t?

Or do I have to use an equation which includes the position as well?
Sorry if my equations are littlebit hard to read (I don´t know the codes to get them look "normal")

I hope some of you might help me out! Would love some tips for b and c as well.

Last edited: Sep 6, 2016
2. Sep 6, 2016

### jbriggs444

Does this remain true if, for instance, the starting separation between the vehicles was 1 meter? Can you come up with a different argument to support the claim that the tractor will stop prior to impact?

3. Sep 6, 2016

### ChrisBrandsborg

Not if the separation is 1 meter, but the separation is 60m.
But I understand what you mean. Maybe it´s more clear to say:

since vA<vB, the tractor will stop before the Tesla will.

4. Sep 6, 2016

### haruspex

Yes. But since you defined them both as positive in the same direction, you mean |vA|<|vB|.

5. Sep 6, 2016

### ChrisBrandsborg

But more importantly, do you know how to solve the problem? How do I find the relative speed on impact?

6. Sep 6, 2016

### haruspex

That gives you the time at which they will meet assuming each has constant acceleration until then. Is that necessarily the case?

7. Sep 6, 2016

### SammyS

Staff Emeritus
You know that the collision happens at t ≈ 3.6 s .

What is the Tesla's speed at that time ?

8. Sep 6, 2016

### ChrisBrandsborg

I have already done that, and found out that the time t = 3.6.
Then I inserted t=3.6 into xB(t) to find the position 12.5m.

What now?

9. Sep 6, 2016

### SammyS

Staff Emeritus
I asked you for the Tesla's speed (velocity will do) NOT its position.

10. Sep 6, 2016

### ChrisBrandsborg

I responded to someone else, but the speed is what I want to find, but I am not sure how. Can I use the equation: v = vi + at?

Or do I have to use an equation which also includes position, since we just found the position (of the crash)

11. Sep 6, 2016

### SammyS

Staff Emeritus
Why not use that?

You used the time to find position. Right?

12. Sep 6, 2016

### ChrisBrandsborg

So, v = 22m/s -5*(3.6) = 4m/s
Yes:)

13. Sep 6, 2016

### SammyS

Staff Emeritus
Of course, you can use equation #2, but it's more involved and should give an equivalent answer.

14. Sep 6, 2016

### jbriggs444

Right. That information is helpful -- it eliminates at least one of the possible sequences of events. But you still need to figure out whether the tractor would stop before or after the collision, if any. @haruspex has hinted at that.

15. Sep 6, 2016

### haruspex

Actually, it is calculating the tractor's speed at that time that is the more revealing.
You don't seem to have understood my post. The way you calculated that time assumed that the accelerations are constant until that time. You need to check whether that is true. See my response to SammyS above.

16. Sep 7, 2016

### ChrisBrandsborg

It is stated in the problem that the acceleration is constant (in the opposite direction of their motion).

"A huge tractor and a Tesla full of school children come driving along a winding mountain road, in opposite directions. The tractor has a speed of 40.0km/h and the car zooms along with 80.0 km/h. The Tesla suddenly comes around a corner, sees the tractor, and they both immediately start braking, both with constant accelerations of 5.00 m/s2 (opposite to their directions of motion)."

17. Sep 7, 2016

### BvU

Acceleration is constant until v = 0, after that moment, it is zero (i.e. a different value!)

18. Sep 7, 2016

### Lis

Can anybody explain how to solve c?

19. Sep 7, 2016

### jbriggs444

If the Tesla is to avoid being hit by the tractor, it must come to a stop at a position the tractor will never reach. How far does the tractor go before stopping in case c?

20. Sep 7, 2016

### BvU

That's not what the problem statement says. Insurance-wise it's always good to stand still when being hit, so the question is valid.

21. Sep 7, 2016

### ChrisBrandsborg

I could also use some help on solving b and c.

22. Sep 7, 2016

### ChrisBrandsborg

On b for instance, do I just check how far both of them drive in the interval of 0.5s, and then afterwards use the same equations as problem a, just change the distance between them?

23. Sep 7, 2016

### BvU

What about a? All done and accounted for ? Result ?

24. Sep 7, 2016

### BvU

Excellent idea. (So: Why ask ?)

25. Sep 7, 2016

### SammyS

Staff Emeritus
Beside the fact that this answer has quite a bit of round-off error ...

You have solved this by assuming that the tractor comes to a stop prior to the collision. You have not confirmed that fact. It's complicated to show that this is the case as a first step, but you can confirm that your results support that view.

In part b, it remains the case that | vA | < | vB |. However, in this case, both vehicles are moving at the time of the collision.

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