Tough distance formula. (Answer provided), but how to get the answer?

  • #1

Homework Statement




A car and a train move together along straight, parallel paths with the same constant cruising speed v[itex]_{0}[/itex] . At t = 0 the car driver notices a red light ahead and slows down with constant acceleration -a[itex]_{0}[/itex]. Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v[itex]_{0}[/itex] with constant acceleration a[itex]_{a}[/itex]. During the same time interval, the train continues to travel at the constant speed v[itex]_{0}[/itex] .


What is the distance traveled by the train during the entire period of (negative and positive) acceleration of the car? Expressed in terms of v[itex]_{0}[/itex] and a[itex]_{0}[/itex] .

I already know the answer is d[itex]_{train}[/itex] = [itex]\frac{2(v_{0}^{2})}{a_{0}}[/itex]

However, I'm still trying to figure out why that is the answer.




Homework Equations



Contant Acceleration Forumula: I believe the relevant equation would be v[itex]_{x^{2}}[/itex]=v[itex]_{0^{2}}[/itex]+2a[itex]_{0}[/itex](x-x[itex]_{0})[/itex]

The Attempt at a Solution



So far I have total time car was accelerating/decelerating: [itex]\frac{v_{0}}{a_{0}}[/itex] + [itex]\frac{v_{0}}{a_{0}}[/itex] = [itex]\frac{2v_{0}}{a_{0}}[/itex]

Now, if I could just figure out how to get this formula: d[itex]_{train}[/itex] = [itex]\frac{2(v_{0}^{2})}{a_{0}}[/itex]
 

Answers and Replies

  • #2
Simon Bridge
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Forget the formulas - draw the velocity time graph.
The area under the graph is the displacement.

The trains graph is flat - so it's distance is the area of a rectangle height v0 and length T. You know how to do this.

The car's v-t graph is a V-shape ... so it must travel half the distance of the train.

The trick it to find the time interval ... the acceleration is the slope of the v-t graph.
Since the whole interval is T, you know the car slows to a stop in T/2 and you know how to find the slope of a graph.
 
  • #3
Unfortunately, I can't forget the formulas because this h/w assignment specifically states that I use the formulas and that I just create a formula expressed in v(sub0) and a(sub0). However, thanks, I appreciate your response on using a probably more logical method and will definately look into the method you describe after I can understand and answer it the only way i'll get credit for on the h/w. It must be possible to derive the formula I provided based on the constant acceleration formula provided, right? Or is this not the correct formula to use? The t variable can't be part of the final equation, so it's gotta be this formula, i think. But how to use that formula to get the answer i provided is where i'm stuck.
 
  • #4
Simon Bridge
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You mean - you must use the kinematic equations directly?

for the train use d=(u+v)T/2
for the car use d=uT+aT2/2

this gives you two equations and two unknowns.

Hint: draw the v-t graph anyway - it tells you how to apply the equations.
 
  • #5
ehild
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What is the distance traveled by the train during the entire period of (negative and positive) acceleration of the car? Expressed in terms of v[itex]_{0}[/itex] and a[itex]_{0}[/itex] .

So far I have total time car was accelerating/decelerating: [itex]\frac{v_{0}}{a_{0}}[/itex] + [itex]\frac{v_{0}}{a_{0}}[/itex] = [itex]\frac{2v_{0}}{a_{0}}[/itex]

Now, if I could just figure out how to get this formula: d[itex]_{train}[/itex] = [itex]\frac{2(v_{0}^{2})}{a_{0}}[/itex]


You got the time: [itex]t=\frac{2v_{0}}{a_{0}}[/itex]
The train travelled with the constant speed v0 during this time. What distance did it travel?


ehild
 
  • #6
You mean - you must use the kinematic equations directly?

for the train use d=(u+v)T/2
for the car use d=uT+aT2/2

this gives you two equations and two unknowns.

Hint: draw the v-t graph anyway - it tells you how to apply the equations.

I'm sorry, but i'm still not sure we are on the same page here. As I said, the answer is already given. I am supposed to express the formula in v[itex]_{0}[/itex] and a[itex]_{0}[/itex]. The equations you are constructing include T's in them which is not what the directions for this particular assignment is calling for. I can't have any variables other than v[itex]_{0}[/itex] and a[itex]_{0}[/itex]. However, thanks for attempting to help me on this, it is appreciated.



You got the time: [itex]t=\frac{2v_{0}}{a_{0}}[/itex]
The train travelled with the constant speed v0 during this time. What distance did it travel?


ehild


Okay, I think I see what you are saying. Distance = velocity * time, thus:

Distance = velocity * [itex]\frac{2v_{0}}{a_{0}}[/itex] = [itex]\frac{2v_{0}^{2}}{a_{0}}[/itex]

Did I follow correctly?

If that is correct, then why doesn't the distance of the car have the same formula? The answer for the total distance of the car is:
[itex]\frac{v_{0}^{2}}{a_{0}}[/itex]

Is that because in order to make this distance[itex]_{car}[/itex] formula, you already know, from the scenario provided, that acceleration (although in opposite directions) is constant value throughout the car's motion, and thus, must come to a conclusion that you can "pretend" the acceleration is in the same direction throughout the entire motion of the car, thus somehow concluding that using only half the car's time, t = [itex]\frac{v_{0}}{a_{0}}[/itex] so that:

Distance = v * t = v * [itex]\frac{v_{0}}{a_{0}}[/itex] = [itex]\frac{v_{0}^{2}}{a_{0}}[/itex]

But how/why am I to know to use only half the total time to produce the car's distance?
 
Last edited:
  • #7
ehild
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Okay, I think I see what you are saying. Distance = velocity * time, thus:

Distance = velocity * [itex]\frac{2v_{0}}{a_{0}}[/itex] = [itex]\frac{2v_{0}^{2}}{a_{0}}[/itex]

Did I follow correctly?

If that is correct, then why doesn't the distance of the car have the same formula? The answer for the total distance of the car is:
[itex]\frac{v_{0}^{2}}{a_{0}}[/itex]
The train moves with constant velocity, the car moves with acceleration. It slows down from v0 to zero velocity, then accelerates back to v0. In average, it moves less fast than the train, the average velocity is just v0/2, so the travelled distance is only half than that of the train.
You certainly learnt some other formula for the displacement of the uniformly accelerating object. For example, this one:

x-x0=0.5(vf+vi)t

ehild
 
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  • #8
Simon Bridge
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I'm sorry, but i'm still not sure we are on the same page here. As I said, the answer is already given. I am supposed to express the formula in v0 and a0. The equations you are constructing include T's in them which is not what the directions for this particular assignment is calling for.
I'm sorry I thought you wanted to know
I already know the answer is [snip]
However, I'm still trying to figure out why that is the answer.
Drawing the v-t diagrams would have told you that.

eg. the car's equation is different due to the areas under the v-t graph - had you drawn them you'd have seen the car's v-t graph has triangles and a triangle is half a rectangle.

The equation I showed you, had T's in them, true, which is why there are two equations. You use one to eliminate T from the other. You substitute your variables into the general formulas and you get the answer ... that way you will have all the answers in terms of the variables that are wanted. If you want separate equations for the train and the car you also need the definition of acceleration: a=(v-u)/T

ehild is working this from the other end which seems to be working better for you :)
this graph + algebra method is more general.
 
  • #9
ehild
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@Simon: LearninDaMath has calculated the time needed to decelerate the car from v0 to full stop and the time of acceleration back with the formula t=v0/a . After that, he used the formula vf2-vi2=2a(x-x0) to get the displacement, which is valid for uniform acceleration. He just forgot the that the train moves with constant velocity.
You are right, Physics has to be understood instead of just using formulae. But lots of schools teach how to find the proper equation and use it from a formula sheet instead of teaching Science.

ehild
 
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  • #10
Simon Bridge
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Yeah I have noticed.
Also there is a feeling among many students that if you are not explicitly told to do it you shouldn't.

I just mistook the question for a different kind of one.
 
  • #11
I hear ya simon. My h/w is not put together by my professor or my school. They use this online system called MasteringPhysics.com. I literally did not have an option to enter any other answer other than the one that ehild guided me towards. Thanks ehild.


By the way, i'm about to post a question that, i believe, does begin to delve into how free fall acceleration relates to a graph. So maybe not all is lost in my school system lol
 
  • #12
Simon Bridge
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I've seen those ... they are six of one and half-a-dozen of the other.
The trick with them is to do the physics first then translate your result into an answer the machine understands. The first part is physics, which is important for teaching/learning and the second is communication - learning is useless if you cannot talk about it and hey: it's marks.
 

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