Solving for Car & Biker's Meeting Point w/ Variable d

[StrawHat]

Homework Statement

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.

Homework Equations

x_final = x_initial + vt
x_final = x_initial + v_initialt + 0.5at²

The Attempt at a Solution

x_bike = -d + v_biket
x_car = 0 + 0 + 0.5at²

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

d = vt - 0.5at²

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

 

[Sleepy_time]

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 x_bike or x_car you should have your answer.

 

[StrawHat]

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 x_bike or x_car you should have your answer.

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 x_bike or x_car and see what I get.

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

I must be doing something wrong again...

 

[Sleepy_time]

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 x_bike or x_car and see what I get.

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

I must be doing something wrong again...

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.

 

[StrawHat]

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.

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:

v > -√(v² - 2ad) and v > √(v² - 2ad)

 

[Sleepy_time]

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

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.

 

[StrawHat]

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.

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

 

[SammyS]

Homework Statement

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.

Homework Equations

x_final = x_initial + vt
x_final = x_initial + v_initialt + 0.5at²

The Attempt at a Solution

x_bike = -d + v_biket
x_car = 0 + 0 + 0.5at²

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

d = vt - 0.5at²

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

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?

 

[Sleepy_time]

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 x_car or x_bike and then use your equation for d to find the distance as some function of d.

 

[StrawHat]

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?

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.

 

[Sleepy_time]

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.

 

[Villyer]

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.

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.

 

[StrawHat]

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 x_car or x_bike and then use your equation for d to find the distance as some function of d.

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:

x_bike = -d + vt

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

 

[Villyer]

Did you get an expression for t?

 

[StrawHat]

Did you get an expression for t?

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

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

x_bike = d = x_car

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

 

[Sleepy_time]

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:

x_bike = -d + vt

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

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 realize how I've come to this.

 

[StrawHat]

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 realize how I've come to this.

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