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

Click For Summary
A collision scenario is presented involving a tractor and a Tesla on a mountain road, with both vehicles starting to brake after a 0.5-second reaction time. The tractor travels at 40 km/h and the Tesla at 80 km/h, both decelerating at 5 m/s². Initial calculations suggest that they collide after 3.6 seconds, with the Tesla reaching a speed of 4 m/s at impact, while the tractor has already stopped. The discussion emphasizes the importance of confirming whether the tractor stops before the collision, as this affects the outcome of the scenario. The conversation highlights the need for careful application of kinematic equations to determine the dynamics of the situation accurately.
  • #31
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).
 
Physics news on Phys.org
  • #32
ChrisBrandsborg said:
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.
 
  • #33
BvU said:
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? :)
 
  • #34
haruspex said:
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.
 
  • #35
ChrisBrandsborg said:
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.
 
  • #36
ChrisBrandsborg said:
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 ?
 
  • #37
BvU said:
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 :)
 
  • #38
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 :sleep: )
 
  • #39
BvU said:
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 :sleep: )

Haha yes :D
Really great to get much help :) Thanks everyone!
 
  • #40
ChrisBrandsborg said:
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 ?
 
  • #41
BvU said:
No, it did not. The 47.65 is 60 m minus the distance the tractor covers till standstill.
Oh. I misread it.
BvU said:
I had thought Haru had gone to bed :sleep: )
no, but it looks like I should have done.
 
  • Like
Likes SammyS
  • #42
haruspex said:
(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 :smile:
 
  • Like
Likes SammyS
  • #43
SammyS said:
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)
 
  • #44
ChrisBrandsborg said:
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?
 
  • #45
ChrisBrandsborg said:
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.
 
  • #46
ChrisBrandsborg said:
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?
 
  • #47
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?
 
  • #48
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? :)
 
  • #49
ChrisBrandsborg said:
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 ?
 
  • #50
BvU said:
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?
 
  • #51
ChrisBrandsborg said:
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?
 
  • #52
oh.. I see.. It should have been 2((-5/2)t^2)
 
  • #53
ChrisBrandsborg said:
the difference between them is now 43.4m
43.3 is a bit closer.
 
  • #54

I re-solved it and got t = 1.78. Correct?
 
  • #55
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.
 
  • #56
BvU said:
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 :)
 
  • #57
ChrisBrandsborg said:
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 :smile:
 
  • #58
BvU said:
My reading for c is that the tractor does not brake, so I don't understand the 2.22222222222222 :smile:

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.
 
  • #59
But in part c, the point is that the tractor does not brake and hence does not stop !
 
  • #60
BvU said:
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
 

Similar threads

Motion in 1-D Problem: Tractor vs Tesla Collision Calculation
  • · Replies 1 ·
Replies
1
Views
1K
Two vehicles braking to avoid a collision
  • · Replies 4 ·
Replies
4
Views
2K
Need quick help with finding positional equation
  • · Replies 1 ·
Replies
1
Views
1K
Help with question on motion: Avoiding a rear-end collision
  • · Replies 6 ·
Replies
6
Views
3K
Acceleration required for ship to avoid crashing into another ship
  • · Replies 15 ·
Replies
15
Views
2K
What Are the Velocities of Two Blocks After an Elastic Collision?
  • · Replies 3 ·
Replies
3
Views
2K
Conservation of momentum question - elastic collisions
  • · Replies 6 ·
Replies
6
Views
5K
Determining position of an object after inelastic collision
  • · Replies 1 ·
Replies
1
Views
2K
Train collision prevention problem
  • · Replies 8 ·
Replies
8
Views
3K
Can You Avoid Hitting the Tractor on a Country Road?
  • · Replies 1 ·
Replies
1
Views
3K