# Finding the distance between an object of const. vel. and another obj. of const. acc.

1. Jun 2, 2012

### StrawHat

1. The problem statement, all variables and given/known data
A car is stopped at an intersection with a red light and a biker (in a bike lane) with velocity v is approaching the car. When the biker is a distance d from the intersection, the light turns green and the car begins to accelerate at a constant acceleration a. If the biker just barely catches up to the car as the car accelerates, how far from the intersection will the biker just catch up to the car? Put your answers only using variable d, not v and a.

2. Relevant equations
xfinal = xinitial + vt
xfinal = xinitial + vinitialt + 0.5at2

3. The attempt at a solution
xbike = -d + vbiket
xcar = 0 + 0 + 0.5at2

Because they should meet at the same location...
xbike = xcar
-d + vt = 0.5at2

d = vt - 0.5at2

But the solution still has v and a, so I am completely stuck.

Last edited: Jun 2, 2012
2. Jun 2, 2012

### Sleepy_time

Re: Finding the distance between an object of const. vel. and another obj. of const.

Hi StrawHat. Your final equation is a quadratic in t. If you solve this for t and use the condition that the biker and the car are at the same position for only a single time, make it so that the answer for t only has real solution. From this you should be able to get an equation for the time and for the distance d. Substituting these back into either xbike or xcar you should have your answer.

3. Jun 2, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

I saw the quadratic too, so I tried to solve for t, and this is what I got:

t = (v ± √(v2 - 2ad)) / a

I will try to substitute this into either xbike or xcar and see what I get.

EDIT: I got xbike = -d + (v2 ± v√(v2 - 2ad))/a

I must be doing something wrong again...

4. Jun 2, 2012

### Sleepy_time

Re: Finding the distance between an object of const. vel. and another obj. of const.

It tells you that the bike barely catches up with the car, i.e. they are at the same position at only one point in time. in your equation you have 2 point because of the ±. So make it so that the ± "vanishes" so you only have one solution for the time.

5. Jun 2, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

How do you tell which one gives you the positive d if v, a, and d are all unknowns?

EDIT: I've tried checking to see which one will give me a positive d, and I got the two equations:

6. Jun 2, 2012

### Sleepy_time

Re: Finding the distance between an object of const. vel. and another obj. of const.

So what you have is:
$t=\frac{v}{a}\pm\sqrt{\frac{v^2}{a^2}-\frac{2d}{a}}$​

You have to have only one solution, not choose either + or -, instead choose what $\frac{v^2}{a^2}-\frac{2d}{a}$ is, so that there is only.

Also, the values for t will both be positive because, you can see that $\frac{v^2}{a^2}\geq\frac{v^2}{a^2}-\frac{2d}{a}$. But those don't matter because if we choose the -ve, then they meet at this time and also at another time for the +ve, so they met twice, which is not what the question wants.

Hope this helps.

7. Jun 2, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

Sorry, I'm still totally stuck. I'm not sure how to get only one solution for the above equation.

8. Jun 2, 2012

### SammyS

Staff Emeritus
Re: Finding the distance between an object of const. vel. and another obj. of const.

Hello StrawHat. Welcome to PF !

If the biker just barely catches up to the car as the car accelerates, then at the point, and time, at which the biker catches up to the car, don't they both have the same velocity?

9. Jun 2, 2012

### Sleepy_time

Re: Finding the distance between an object of const. vel. and another obj. of const.

Ok, so to get one solution choose that $\frac{v^2}{a^2}-\frac{2d}{a}=0$, this is so that we get rid of the ± which gives 2 solutions. If you input this into the equation for time, as well as re-arrange for d, you should have d and t as functions of a and v. Substitute back into xcar or xbike and then use your equation for d to find the distance as some function of d.

10. Jun 2, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

I don't think so. The bike is traveling at a constant velocity from a distance of d, so if the car was going at the same velocity, the bike would never catch up to the car with a constant acceleration.

11. Jun 2, 2012

### Sleepy_time

Re: Finding the distance between an object of const. vel. and another obj. of const.

If you're not convinced by what I did, you can also differentiate your function of time with respect to d and equal it to infinity (or re-arrange for d=f(t) and differentiate this with respect to time and equal it to 0 because $\frac{d(d)}{dt}=(\frac{d(t)}{dd})^{-1}=0$) this is to find the maximum/minimum values of d for any time. We want the maximum value because the equation is for when they meet, just we want the furthest so that they just meet. This will give the same answer.

Last edited: Jun 2, 2012
12. Jun 2, 2012

### Villyer

Re: Finding the distance between an object of const. vel. and another obj. of const.

But the problem says that the biker just barely catches up to the car.
This means that the biker never passes the car, he just momentarily is at the same position when the car again overtakes the biker.

13. Jun 2, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

After I did $\frac{v^2}{a^2}-\frac{2d}{a}=0$, this is what I ended up with:

$d = \frac{v^2}{2a}$

I plugged it into this:

xbike = -d + vt

And I still can't manage to get an answer with only d.

14. Jun 2, 2012

### Villyer

Re: Finding the distance between an object of const. vel. and another obj. of const.

Did you get an expression for t?

15. Jun 2, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

Yes, $t=\frac{v}{a}$

EDIT: Further plugging and chugging led to an answer of this:

xbike = d = xcar

I guess that's the right answer...?

Last edited: Jun 2, 2012
16. Jun 2, 2012

### Sleepy_time

Re: Finding the distance between an object of const. vel. and another obj. of const.

Ok so:
$x_{bike}=-d+v\frac{v}{a}=-d+\frac{v^2}{a}$

You then have $d=\frac{v^2}{2a} \rightarrow x=-d+2d=d$, you chose your coordinates such that the intersection is at x=0, so they'll meet a distance d after the intersection.

I hope you realise how I've come to this.

17. Jun 3, 2012

### StrawHat

Re: Finding the distance between an object of const. vel. and another obj. of const.

Yes, and that is how I got the answer d. Thank you so much for your help!