Will the Tractor and Tesla Avoid a Collision on the Mountain Road?

[ChrisBrandsborg]

Homework Statement

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))?

Homework 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.

 

[jbriggs444]

Since vA < vB, we know that the tractor will stop before the Tesla will hit it.

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?

 

[ChrisBrandsborg]

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?

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.

 

[haruspex]

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.

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

 

[ChrisBrandsborg]

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

 

[haruspex]

Then I set xA = xB(t)

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

 

[SammyS]

...

3. The Attempt at a Solution :
...

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?

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

What is the Tesla's speed at that time ?

 

[ChrisBrandsborg]

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

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?

 

[SammyS]

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?

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

 

[ChrisBrandsborg]

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

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)

 

[SammyS]

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?

Why not use that?

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

You used the time to find position. Right?

 

[ChrisBrandsborg]

Why not use that?

You used the time to find position. Right?

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

 

[SammyS]

Yes

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

 

[jbriggs444]

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.

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.

 

[haruspex]

What is the Tesla's speed at that time ?

Actually, it is calculating the tractor's speed at that time that is the more revealing.

already done that, and found out that the time t = 3.6.

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.

 

[ChrisBrandsborg]

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.

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)."

 

[BvU]

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

 

[Lis]

Can anybody explain how to solve c?

 

[jbriggs444]

Can anybody explain how to solve c?

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?

 

[BvU]

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?

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.

 

[ChrisBrandsborg]

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

 

[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?

 

[BvU]

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

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

 

[BvU]

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?

Excellent idea. (So: Why ask ?)

 

[SammyS]

Homework Statement

Problem 5: Collision course
...

Homework 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

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
...

Maybe it´s more clear to say:

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

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

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 | v_A | < | v_B |. However, in this case, both vehicles are moving at the time of the collision.

 

[BvU]

Excellent idea. (So: Why ask ?)

However, if that shows there IS a collision, things change, of course. Hence my interest in the result of a.
[edit] or was there a collision in a also ?

 

[jbriggs444]

[with respect to part c]

That's not what the problem statement says.

to have time to stop before being hit by the tractor

Right you are, thank you for the correction. So the Tesla must come to a stop exactly at a point where the tractor will be, at the exact moment when the tractor arrives there.

 

[ChrisBrandsborg]

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

Yes, at least I think so :)

After I found that t must be = 3.6, I put that into the xB(t) function to find the position of the crash.
I found that xB = 12.5m (this is the position of the crash -> 12.5m from the starting point of the tractor)
-> Which mean that the Tesla drove 60m - 12.5m = 47.5m before they crashed.

Then I used the equation: Vhit = Vb + at

Vhit = -22.2m/s + 5.5m/s^2 * (3.6s) = -4m/s (which means that Tesla had a velocity of 4m/s in negative direction)

 

[ChrisBrandsborg]

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 | v_A | < | v_B |. However, in this case, both vehicles are moving at the time of the collision.

How do I check if the tractor comes to a spot prior to the collision?

 

[ChrisBrandsborg]

However, if that shows there IS a collision, things change, of course. Hence my interest in the result of a.
[edit] or was there a collision in a also ?

Yes, there was a collision in a. 3.6 seconds after they started braking. The Tesla had then a velocity of 4m/s when they crashed and the tractor had already stopped.

 

[BvU]

I did the math for a in the mean time and agree with #28.
so my # 24 is slightly hasty: first have to check if the tractor is or is not moving when hit (but that will come out when hit time is evaluated).

 

[haruspex]

Yes, there was a collision in a. 3.6 seconds after they started braking. The Tesla had then a velocity of 4m/s when they crashed and the tractor had already stopped.

But your 3.6 seconds calculation assumed they had constant acceleration until that time, no? If the tractor had stopped it no longer had that acceleration.

 

[ChrisBrandsborg]

I did the math for a in the mean time and agree with #28.
so my # 24 is slightly hasty: first have to check if the tractor is or is not moving when hit (but that will come out when hit time is evaluated).

So how do you check if the tractor is moving? Can you write an equation and solve it? :)

 

[BvU]

But your 3.6 seconds calculation assumed they had constant acceleration until that time, no? If the tractor had stopped it no longer had that acceleration.

No, it did not. The 47.65 is 60 m minus the distance the tractor covers till standstill.

 

[jbriggs444]

How do I check if the tractor comes to a spot prior to the collision?

Case by case analysis is the obvious way.
Suppose that the tractor comes to a stop and see whether the Tesla has passed the position of the tractor's stopping prior to the moment of the tractor's stopping.