1. ## Word Problem

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 36 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)

No clue how to do it.

2. Originally Posted by Esthephane
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 36 ft/s2. What is the distance covered before the car comes to a stop? (Give your answers correct to one decimal place.)

No clue how to do it.
Use the kinematic equations

$x=x_{0}+v_{0}t+\frac{1}{2}at^2$

and

$v=v_{0}+at$

Dont forget to convert to the same units!

3. Originally Posted by TheEmptySet
Use the kinematic equations

$x=x_{0}+v_{0}t+\frac{1}{2}at^2$

and

$v=v_{0}+at$

Dont forget to convert to the same units!

I'm trying to figure out how to use the equations you gave me but I don't understand how. I'm looking at my book and i see some similar problems but doesn't help a lot. Think you could explain a little further?

4. Originally Posted by Esthephane
I'm trying to figure out how to use the equations you gave me but I don't understand how. I'm looking at my book and i see some similar problems but doesn't help a lot. Think you could explain a little further?
To use the 2nd equation you need to know the inital and final velocity and the acceleration to find the time it takes to stop.

You are given explicity the initial velocity $v_0$ and the acceleration $a$. But we want to solve for the time $t$ when the car is stopped so we know that $v=0$

Using this you can solve for the time $t$ it takes for the car to stop.

After you get this information you can use all of the previous plus the new information to solve for $x$ the stopping distance. Also note for convience you can assume that $x_0=0$.

5. Originally Posted by TheEmptySet
To use the 2nd equation you need to know the inital and final velocity and the acceleration to find the time it takes to stop.

You are given explicity the initial velocity $v_0$ and the acceleration $a$. But we want to solve for the time $t$ when the car is stopped so we know that $v=0$

Using this you can solve for the time $t$ it takes for the car to stop.

After you get this information you can use all of the previous plus the new information to solve for $x$ the stopping distance. Also note for convience you can assume that $x_0=0$.
Great, It's making more sense to me than before. But, I have one last question. I'm trying to convert the units and what i have for t is t=-50/36, how do i convert the units.

6. I think its better to understand those formulas. Have you ever asked yourself where those formulas came from?

You do not need formulas for this problem, just a little bit logic.

First 1 foot = 0.000189(39) miles. So 1 mile is 1/0.000189(39)=5280 ft

Now 1 hour = 60 min.
1 min = 60 sec. so 60 min = 60*60sec=3600sec.

So the speed that you are traveling with is 50*5280ft/3600s = 73,(3) ft/s

Now you have a constant deceleration of 36 ft/s2.

What that really means? That means that the speed drops for 36 units every second.

Now how many seconds do we need so that the speed will drop to 0?

Now do some ratio.

1s____x s (seconds)
----=----
36____73 (drop of units ft/s)

(we can do this because of constant deacelleration)

Now just find x and tell me how many seconds does the car need so that it will stop?

You will tell me the feet that the car has covered.