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

[BvU]

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

What do you get for the collision time when doing b the same way as a ?

 

[ChrisBrandsborg]

What do you get for the collision time when doing b the same way as a ?

I haven´t done it yet. I will solve it later today! I have some other stuff to do! I will post it here when I have tried solving b and c :)

 

[BvU]

Chris is a lucky guy: he gets so much help it's become dazzling. Must be a really nice exercise ! (I only barged in because I had thought Haru had gone to bed )

 

[ChrisBrandsborg]

Chris is a lucky guy: he gets so much help it's become dazzling. Must be a really nice exercise ! (I only barged in because I had thought Haru had gone to bed )

Haha yes :D
Really great to get much help :) Thanks everyone!

 

[SammyS]

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

Where does the tractor come to a stop?

Where is the Tesla at this time? How far has it traveled ?

 

[haruspex]

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

Oh. I misread it.

I had thought Haru had gone to bed )

no, but it looks like I should have done.

 

[BvU]

(re sleeping): no, but it looks like I should have done.

Ah, turns out it's only 22:30 in Sydney, so I was still in my after-lunch dip

 

[ChrisBrandsborg]

Where does the tractor come to a stop?

Where is the Tesla at this time? How far has it traveled ?

Okay, I used v = vi + at
0 = 11.1 m/s -5t
5t = 11.1

t = (11.1/5) = 2.22 (which means that the tractor will stop at t = 2.22) --> When t = 2.22 the Tesla is still moving (since we have already solved that he will hit the tractor at t = 3.6)

 

[ChrisBrandsborg]

I haven´t done it yet. I will solve it later today! I have some other stuff to do! I will post it here when I have tried solving b and c :)

I solved first that the tractor will stop 2.2 seconds after he start braking. Then when I used the 0.5s delay thing, I solved that they will crash at t = 1.74, which means that the tractor also has a velocity when they crash.

But is that correct? Can I use my equations when Vtractor(final) isn't = 0 ?
I have used the "timeless" equation, but I have put vfinal = 0 to find the collision-time.. hmm?

 

[BvU]

I solved first that the tractor will stop 2.2 seconds after he start braking. Then when I used the 0.5s delay thing, I solved that they will crash at t = 1.74, which means that the tractor also has a velocity when they crash.

But is that correct?

Yes, so now you can't solve for the Tesla solo as in case a.

Can I use my equations when Vtractor(final) isn't = 0 ?
I have used the "timeless" equation, but I have put vfinal = 0 to find the collision-time.. hmm?

Both have to be in the same place at the same time. Write down the equations for x (distance from each other) = 0 . vfinal is no longer zero for either, but if you have t you can find them easily.

 

[SammyS]

Okay, I used v = vi + at
0 = 11.1 m/s -5t
5t = 11.1

t = (11.1/5) = 2.22 (which means that the tractor will stop at t = 2.22) --> When t = 2.22 the Tesla is still moving (since we have already solved that he will hit the tractor at t = 3.6)

The Tesla is still moving at t = 2.22, but where is it? Has it gotten far enough to have collided with the tractor?

 

[ChrisBrandsborg]

I have tried to solve b, this is what I got (dont know if its correct):

vA = 11.1m/s, which means in 0.5s he will drive 5.55 = 5.6m
vB = -22.2m/s which means in 0.5s he will drive -11.1m (so the difference between them is now 43.4m instead of 60)

Then I used the same equation as earlier to find if they hit before/after the tractor stops. I found out that t = 1.74, which means before the tractor stops,
which also mean that I have to use another equation, since I have used vAfinal = 0.

So now I have these equations instead for position:
xA(t) = (-5/2)t^2 + 11.1t
xB(t) = 5/2t^2 -22.2t + 43.4

Then I put xA(t) = xB(t) to find out what t is when they are in the same position.

--> t = 1.06

Then I put xB(1.06) to find out the position of the Tesla when they crash. Which was 22.7m (from the initial position)

Then I found the velocity on impact:

vAfinal = 11.1m/s - 5.5*1.06 = 5.8m/s = 21km/h
vBfinal = -22.2m/s + 5.5*1.06 = 16.9m/s = 61km/h

Is that correct? Are these velocities too fast?

 

[ChrisBrandsborg]

I tried c as well:

I used the tA-stop = 2.2, to find out how many meters the tractor drove before he stopped, and then added that up with the meters both of them drove before they started braking. I got 29m (which means that the remaining distance is 31m)

so I know that the Tesla has to stop before 31m.
I put vBfinal = 0:

a = (vBfinal^2 - vBinitial^2)/2x
a = 493/62 = 7.951 = 8m/s^2 (since I have been using 1 decimal only from the start)

That´s looks correct? Comments? :)

 

[BvU]

So now I have these equations instead for position:
xA(t) = (-5/2)t^2 + 11.1t
xB(t) = 5/2t^2 -22.2t + 43.4

Then I put xA(t) = xB(t) to find out what t is when they are in the same position.

--> t = 1.06

I don't see the 1.06 seconds satisfy xA = xB ?

 

[ChrisBrandsborg]

I don't see the 1.06 seconds satisfy xA = xB ?

No? hmm..

(-5/2)t^2 + 11.1t = (5/2)t^2 - 22.2t + 43.4

((-5/2)t^2)^2 -33.3t - 43.4 = 0

(25/4)t^4 - 33.3t - 43.4 = 0

I soved it with GeoGebra and got t = 1.06 or t = some negative number (which is not valid)
Is it wrong?

 

[BvU]

No? hmm..

(-5/2)t^2 + 11.1t = (5/2)t^2 - 22.2t + 43.4

((-5/2)t^2)^2 -33.3t - 43.4 = 0

I soved it with GeoGebra and got t = 1.06 or t = some negative number (which is not valid)
Is it wrong?

 

[ChrisBrandsborg]

oh.. I see.. It should have been 2((-5/2)t^2)

 

[haruspex]

the difference between them is now 43.4m

43.3 is a bit closer.

 

[ChrisBrandsborg]

I re-solved it and got t = 1.78. Correct?

 

[BvU]

Within 1%. But you shouldn't have to ask because you can check for yourself, right?
Backing up Haru: generally, work with symbols as long as feasible, then work with plenty digits and round off at the very last.

 

[ChrisBrandsborg]

Within 1%. But you shouldn't have to ask because you can check for yourself, right?
Backing up Haru: generally, work with symbols as long as feasible, then work with plenty digits and dound off at the very last.

Yes :)
Yeah, I will do from now, but since I have already done so much with 1 decimal, then I´ll just stick with it for this problem.
Thanks a lot :)

 

[BvU]

I tried c as well:

I used the tA-stop = 2.2, to find out how many meters the tractor drove before he stopped, and then added that up with the meters both of them drove before they started braking. I got 29m (which means that the remaining distance is 31m)

so I know that the Tesla has to stop before 31m.
I put vBfinal = 0:

a = (vBfinal^2 - vBinitial^2)/2x
a = 493/62 = 7.951 = 8m/s^2 (since I have been using 1 decimal only from the start)

That´s looks correct? Comments? :)

My reading for c is that the tractor does not brake, so I don't understand the 2.22222222222222

 

[ChrisBrandsborg]

My reading for c is that the tractor does not brake, so I don't understand the 2.22222222222222

To find where the tractor stops I used:

vAfinal = vAinitial + at
0 = 11.1m/s -5t
5t = 11.1m/s
t = 11.1/5 = 2.2222222 :)

In problem c I just used the t = 2.2 to find out how many meters he drove before he stopped, so that I know how much the distance was from A to B.
So that I could solve the aB (acc.) Remaining distance: 31m (x)

so aB (acc.) = vBinitial^2 / 2*31m --> 493/62 = 7.951 = approximately 8m/s^2 (so that is the acceleration for the Tesla for him to stop before he hits the Tractor) if we set his velocity direction to be negative. The acceleration is the opposite direction of the velocity direction.

 

[BvU]

But in part c, the point is that the tractor does not brake and hence does not stop !

 

[ChrisBrandsborg]

But in part c, the point is that the tractor does not brake and hence does not stop !

The Tractor does brake? if not they would crash no matter what :P
He brakes after 0.5s (same as b) with accelration -5m/s^2

 

[jbriggs444]

The Tractor does brake? if not they would crash no matter what :P
He brakes after 0.5s (same as b) with accelration -5m/s^2

The layout of the problem statement may be confusing. It take the form of a scenario description and a question (question a), a modification to the scenario and another question (question b) and a final modification to the scenario and a final question (question c).

The scenario in question c is not described after the "c)", but before it.

[emphasis mine]

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))?

 

[ChrisBrandsborg]

The layout of the problem statement may be confusing. It take the form of a scenario description and a question (question a), a modification to the scenario and another question (question b) and a final modification to the scenario and a final question (question c).

The scenario in question c is not described after the "c)", but before it.

[emphasis mine]

Oh.. I didnt read the "sentence" over "c)"...
But don´t they crash no matter what if he continues with his original speed without braking?

 

[jbriggs444]

But don´t they crash no matter what if he continues with his original speed without braking?

They crash, yes. But that's not what question c is asking.

 

[ChrisBrandsborg]

They crash, yes. But that's not what question c is asking.

Oh, yeah, I understand.. They will crash, but the Tesla will have a velocity = 0 when the Tractor crash into him.

 

[BvU]

The reason you didn't read it in the papers is that in fact this winding mountain road was more than wide enough for them to pass by each other
But for the exercise the composer really meant to type the words 'single lane' and 'narrow'.

Now, equations, if you please ...

 

[ChrisBrandsborg]

Okay, so how do we solve c:

First find out the remaining distance after the 0.5s delay:
We already have found that out: 43.3m

Now what?

xB(t) = 1/2at^2 - 22.2t + 43.3

and we know that the tractor has vA constant = 11.1m/s
I need to find t or x, to find a. How?

 

[jbriggs444]

You are trying for a situation where the tractor's position and the car's position coincide at the same time and the same place that the car comes to a stop. [Brake any harder and you've wasted braking force and are waiting for the tractor to arrive. Brake any softer and you aren't stopped when the crash occurs].

That's two conditions you are trying to fit simultaneously -- a time and a place. Maybe you can write two simultaneous equations, one for each of the two conditions.

 

[ChrisBrandsborg]

You are trying for a situation where the tractor's position and the car's position coincide at the same time and the same place that the car comes to a stop. [Brake any harder and you've wasted braking force and are waiting for the tractor to arrive. Brake any softer and you aren't stopped when the crash occurs].

That's two conditions you are trying to fit simultaneously -- a time and a place. Maybe you can write two simultaneous equations, one for each of the two conditions.

Can you give me another hint :-) What equation I can use?

 

[jbriggs444]

Can you give me another hint :-) What equation I can use?

Can you write down a formula for the tractor's position as a function of time?
Can you write down a formula for the Tesla's position as a function of time? (keep everything symbolic, including its acceleration)..
At t=T_c, the time of the collision, what must be true of these two functions?

 

[ChrisBrandsborg]

Can you write down a formula for the tractor's position as a function of time?
Can you write down a formula for the Tesla's position as a function of time? (keep everything symbolic, including its acceleration)..
At t=T_c, the time of the collision, what must be true of these two functions?

Symbolic functions:
xA(t) = (aA/2)t^2 + vA(initial)*t
xB(t) = (aB/2)t^2 + vB(initial)*t + xBi (distance)

With numbers:
xA(t) = 11(m/s)*t
xB(t) = (5/2)t^2 - 22.2t + 43.3

at t=Tc, they must be at the same position, so I can put xA(t) = xB(t)?

vA(initial)*t = (aB/2)t^2 + vB(initial)*t + xBi (distance)

=> (aB/2)t^2 + vB(initial)*t - vA(initial)*t + xBi = 0

With numbers:

aA = 0, so (aA/2)t^2 goes away:

(5/2)t^2 + 22.2t - 11.1t + 43.3 = 0

(5/2)t^2 + 11.1t + 43.3 = 0

I don't get any solutions for this one, but if the distance is negative?

-43.3, then I get t = 2.50.

Is that valid?