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

[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?

 

[BvU]

1. Tractor does ot brake
2. car does not move at t
3. car does not decelerate with 5 m/s^2
and on top of that: 1.78 can not satisfy your equation for both coefficients of t^2 (5/2 c.q. 10/2).

 

[ChrisBrandsborg]

1. Tractor does ot brake
2. car does not move at t

Can I use the t = 1.78 to find out the acceleration it must have to stop by then:

so a = -vB/t

a = -(-22.2m/s)/1.78 = 12.5m/s^2 ?

 

[BvU]

No. That was the result of a calculation where the tractor braked too.

 

[ChrisBrandsborg]

No. That was the result of a calculation where the tractor braked too.

Yeah, true,

No. That was the result of a calculation where the tractor braked too.

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

Is this correct? I have put tractor´s acceleration = 0.

 

[BvU]

Good. Gives you one equation. But you have two unknowns...

 

[ChrisBrandsborg]

Good. Gives you one equation. But you have two unknowns...

Yes, true... aB and t, so what now ?
I need one more equation, so that I can put one into the other?

 

[BvU]

You haven't made use yet of the fact that the Tesla has come to a halt

 

[ChrisBrandsborg]

vBfinal

You haven't made use yet of the fact that the Tesla has come to a halt

No, so we have also this equation:

Convert vfinal = vi + at, to get t alone:

t = (-vB/a), and this we can insert into the other one, so that we only have aB left?

 

[BvU]

Don't ask that much. If you convince yourself it's the thing to do, go ahead and do !

 

[ChrisBrandsborg]

Don't ask that much. If you convince yourself it's the thing to do, go ahead and do !

Hehe, I do ask too much :D
I´ll try it out, and see if I can solve the puzzle :D

 

[ChrisBrandsborg]

I´ll did the math, and solved that aB = 2.55, rounded to 2.6m/s^2 :)

 

[BvU]

And you convinced yourself ?

Not me, though. If 5 m/s^2 isn't enough with a tractor that brakes, then it surely isn't enough for a tractor that doesn't brake !

 

[ChrisBrandsborg]

And you convinced yourself ?

Not me, though. If 5 m/s^2 isn't enough with a tractor that brakes, then it surely isn't enough for a tractor that doesn't brake !

hmm.. You´re saying something there.. Where did a do wrong?

I put t = (-vB/aB) into the position function:

x(t) = (aB/2)t^2 + vB(t) - vA(t) + xr (distance)

x((-vB/aB)) = (aB/2)(-vB/aB)^2 + vB((-vB/aB)) - vA((-vB/aB)) + xr (distance)

--> (vB^2*aB)/2aB - (vB^2)/aB + (vAvB)/aB + xR

Then I multiplied by 2aB to make it easier:

vB^2*aB - 2vB^2 + 2(vAvB) + 2xR*aB

Then I put in -22.2 for vB, and 43.3 for xR.

Can you see some errors?

 

[jbriggs444]

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

BvU is too nice to say this, but I find the above annoying for several reasons.

First, the variable names are not declared -- you do not say what the variable names mean. Here xA(t) apparently denotes tractor position as a function of t, aA denotes the tractor acceleration, vA(initial) denotes the tractor's initial velocity, ... and xBi (distance) is the initial distance of the Tesla from the tractor.

Second, you use two different conventions for indicating initial values. One is "(initial)" and one is "i (distance)". And you've already adopted a convention where the parentheses enclose the function argument -- numerical time. Consistency helps readability.

Third, formatting could use work. Subscripts and superscripts are easy to do even without TeX.

x_B(t) = a_Bt² + v_B0t + x_B0 reads much more easily than xB(t) = (aB/2)t^2 + vB(initial)*t + xBi (distance)

Edit: My background is as a computer programmer trained in the 70's. We were taught to declare and document all variable names and that computer programs are written as much for the next guy who will maintain them, as for the computer that will run them. That training colors my approach to physics somewhat.

 

[ChrisBrandsborg]

BvU is too nice to say this, but I find the above annoying for several reasons.

First, the variable names are not declared -- you do not say what the variable names mean. Here xA(t) apparently denotes tractor position as a function of t, aA denotes the tractor acceleration, vA(initial) denotes the tractor's initial velocity, ... and xBi (distance) is the initial distance of the Tesla from the tractor.

Second, you use two different conventions for indicating initial values. One is "(initial)" and one is "i (distance)". And you've already adopted a convention where the parentheses enclose the function argument -- numerical time. Consistency helps readability.

Third, formatting could use work. Subscripts and superscripts are easy to do even without TeX.

x_B(t) = a_Bt² + v_B0t + x_B0 reads much more easily than xB(t) = (aB/2)t^2 + vB(initial)*t + xBi (distance)

Sorry for making it hard! I haven't learned the "codes" to make them look nicer, but I will try to learn it! I understand that it is hard to understand what I mean, and hard to read.

 

[jbriggs444]

Sorry for making it hard! I haven't learned the "codes" to make them look nicer, but I will try to learn it! I understand that it is hard to understand what I mean, and hard to read.

There is a bar of icons above the editting pane where you can pull down subscripts (the x₂) icon) and superscripts (the x² icon). This brings the tags you need into your document. You can use preview to see how they render.

Or you can just write the tags yourself.

Regular[sup]superscript[/sup][sub]subscript[/sub] renders as Regular^superscript_subscript

Information on that is in https://www.physicsforums.com/help/bb-codes

If you want to get fancier, you can use TeX like \frac{x^2}{2} which (when enclosed in proper tags) renders as $\frac{x^2}{2}$

See https://www.physicsforums.com/help/latexhelp/

 

[ChrisBrandsborg]

There is a bar of icons above the editting pane where you can pull down subscripts (the x₂) icon) and superscripts (the x² icon). This brings the tags you need into your document. You can use preview to see how they render.

Or you can just write the tags yourself.

Regular[sup]superscript[/sup][sub]subscript[/sub] renders as Regular^superscript_subscript

Information on that is in https://www.physicsforums.com/help/bb-codes

If you want to get fancier, you can use TeX like \frac{x^2}{2} which (when enclosed in proper tags) renders as $\frac{x^2}{2}$

See https://www.physicsforums.com/help/latexhelp/

Cool, thanks! I´ll check it out :-)

 

[BvU]

($\TeX$ commercial:)

It's so easy ! just put $or around the expression to get$LaTeX$. So (aB/2)t^2 + vB(initial)*t - vA(initial)*t + xBi = 0 becomes (aB/2)t^2 + vB(initial)*t - vA(initial)*t + xBi = 0 Then: superscripts come with ^ and subscripts with _ and grouping goes with curly brackets {...} You learn from other posts with a right-click on the formula and picking Show Math as ...$\TeX$commands. - - - - - - - Anyway, we had {1\over 2 } a t^2 + v_{B,i} \; t - v_{A,i}\; t + x_{B,i} = 0 ( I use a instead of aB) and 0 - v_{B,i} = at so what's easier than rewrite the quadratic term as$-{1\over 2 }v_{B,i} \;t $? and then use$t$to get$a##.

 

[ChrisBrandsborg]

Can you find some errors? I will try to make it look nicer, so it is easier to see:
x(t) = position function
v_A = Tractor´s initial speed (constant in c)
v_B = Tesla´s initial speed
t = time after the Tesla start braking
x_r = distance remaining between the cars after the 0.5s delay
a_B = Acceleration of Tesla

I put t = (-v_B/a_B) into the position function:
x(t) = a_B/2)t² + v_B(t) - v_A(t) + x_r
x(-v_B/a_B) = (a_B/2)(-v_B/a_B)² + v_B(-v_B/a_B) - v_A(-v_B/a_B) + x_r

--> (v_B²a_B)/(2a_B) - (v_B²)/(a_B) + (v_Av_B)/(a_B) + x_r

Then I multiplied by 2a_B to make it easier:

--> v_B²a_B - 2v_B² + 2v_Av_B + 2x_ra_B

Then I put in -22.2 for v_B, and 43.3 for x_r.

Can you see some errors?

 

[ChrisBrandsborg]

($\TeX$ commercial:)

It's so easy ! just put $or around the expression to get$LaTeX$. So (aB/2)t^2 + vB(initial)*t - vA(initial)*t + xBi = 0 becomes (aB/2)t^2 + vB(initial)*t - vA(initial)*t + xBi = 0 Then: superscripts come with ^ and subscripts with _ and grouping goes with curly brackets {...} You learn from other posts with a right-click on the formula and picking Show Math as ...$\TeX$commands. - - - - - - - Anyway, we had {1\over 2 } a t^2 + v_{B,i} \; t - v_{A,i}\; t + x_{B,i} = 0 ( I use a instead of aB) and 0 - v_{B,i} = at so what's easier than rewrite the quadratic term as$-{1\over 2 }v_{B,i} \;t $? and then use$t$to get$a##.

How do you get $-{1\over 2 }v_{B,i} \;t$ ?