# Finding the stopping distance of a car? Help please.

## Homework Statement

I am given 3 variables, Vxi (initial velocity), Tr (reaction time), Ax (acceleration.

Now, I must make a formula for the braking distance.

Here are the variables: Vxi = 20 m/s, Tr = 0.60 s, Ax = -4 m/s^2

How would I construct there in a formula to get the total braking distance d = 62 m?

## The Attempt at a Solution

Yes the answer is given, but I do not know the formula used, given the variables I stated above to get d = 62 m?

Any help is much appreciated!

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Do you know the formulas for motion with constant acceleration?

Method 1:

You also know that Vx=0, since the car is going to stop.
Using Vx=Vxi+Ax*Tb, where Tb is braking time, to find Tb, gives you the opportunity to find d using the formula:

d=Vxi *T + (1/2)*Ax*T^2 , where T=Tr+Tb.

Method 2:

Use 2Ax*d=Vx^2-Vxi^2 to find distance traveled during braking, and use d=Vxi*Tr to find distance traveled during the driver's reaction time. Then find d by adding these to displacements together.

Feel free to choose the method you feel most convenient...

Any further questions?

I s hould of mentioned that t time should not appear in the formula.

Do you know the formulas for motion with constant acceleration?

Method 1:

You also know that Vx=0, since the car is going to stop.
Using Vx=Vxi+Ax*Tb, where Tb is braking time, to find Tb, gives you the opportunity to find d using the formula:

d=Vxi *T + (1/2)*Ax*T^2 , where T=Tr+Tb.

Method 2:

Use 2Ax*d=Vx^2-Vxi^2 to find distance traveled during braking, and use d=Vxi*Tr to find distance traveled during the driver's reaction time. Then find d by adding these to displacements together.

Feel free to choose the method you feel most convenient...

Any further questions?

Okey, I think my method 2 is somewhat easier than metod 1.
You must take the reaction time into consideration, because it's the time before the driver pushes the brake in, and Ax starts working, therefore:

Find distance traveled during reaction time:

Dr=Vxi*Tr = 20m/s*0.60s =

Find distance traveled during braking:

$$2Ax *Db=Vx^2-Vxi^2$$ gives

$$Db=\frac {Vx^2-Vxi^2}{2Ax}=\frac {(0 m/s)^2-(20m/s)^2}{2*-4 m/s^2}$$

Stopping distance is then: d=Dr+Db.

This will give you the 62 m you need to have...

Last edited:

Okey, I think my method 2 is somewhat easier than metod 1.
You must take the reaction time into consideration, because it's the time before the driver pushes the brake in, and Ax starts working, therefore:

Find distance traveled during reaction time:

Dr=Vxi*Tr = 20m/s*0.60s =

Find distance traveled during braking:

$$2Ax *Db=\frac {Vx^2-Vxi^2}$$ gives

$$Db=\frac {Vx^2-Vxi^2}{2Ax}=\frac {(0 m/s)^2-(20m/s)^2}{2*-4 m/s^2}$$

Stopping distance is then: d=Dr+Db.

This will give you the 62 m you need to have...

Dr and Db ?What 2 variables are you figuring out, and then subtracting them?

Dr and Db ?What 2 variables are you figuring out, and then subtracting them?
Since there are two different parts of this movement

Dr = distance traveled during reaction time.

During the 0.60 s of the reaction time, the car moves with the initial velocity, Vxi all the time, then the reaction time comes to and end after 0.60 s, when the driver of the car pushes the brake in. Here acceleration is zero, and we can use v=distance/time to find distance= velocity * time.

Db= Distance traveled during braking (that is with the brake pushed in).

When the driver of the car has ushed the brake in, the car starts to decelerate with the acceleration of -4 m/s^2. When the car stops, the speed is 0.
So we can solve the equation I gave you for d, here Db, to find distance traveled after the driver has applied to the brakes.

I hope this explained it somehow, if something's still unclear, please tell me ...

wow, you explained perfect! Thank you mstub

Since there are two different parts of this movement

Dr = distance traveled during reaction time.

During the 0.60 s of the reaction time, the car moves with the initial velocity, Vxi all the time, then the reaction time comes to and end after 0.60 s, when the driver of the car pushes the brake in. Here acceleration is zero, and we can use v=distance/time to find distance= velocity * time.

Db= Distance traveled during braking (that is with the brake pushed in).

When the driver of the car has ushed the brake in, the car starts to decelerate with the acceleration of -4 m/s^2. When the car stops, the speed is 0.
So we can solve the equation I gave you for d, here Db, to find distance traveled after the driver has applied to the brakes.

I hope this explained it somehow, if something's still unclear, please tell me ...

Also I am not subtracting these two, I'm adding them together :)

It's always a good feeling when someone understands what you're trying to explain :D

Hey, having a little trouble here.

I get 12m for Dr

But im not getting correct numbers for Db ? im getting 2.5...?

How do you do that?

$$Db=\frac {Vx^2-Vxi^2}{2Ax}=\frac {(0 m/s)^2-(20m/s)^2}{2*-4 m/s^2}=\frac {400 (m/s)^2}{8m/s^2} = 50$$

Look over your work once again.

Sorry for the late answer, I only got an error message when trying to post: 504 gateway error...

$$2Ax *Db=Vx^2-Vxi^2$$ gives

$$Db=\frac {Vx^2-Vxi^2}{2Ax}=\frac {(0 m/s)^2-(20m/s)^2}{2*-4 m/s^2}$$

Stopping distance is then: d=Dr+Db.

This will give you the 62 m you need to have...

And Dr is correct