# How to solve for the distance traveled given force, mass and speed

## Homework Statement

A car has a mass of 1500kg. If the driver applies the breaks, the maximum amount of friction force that can be applies without skidding is 7000N. If the car is traveling at 25m/s, what is the shortest distance the car can stop safely?

## Homework Equations

I'm assuming it has something to do with F=mα and using that answer to solve for distance using the kinematics equation: Vfinal(squared)=Vinitial+2α[(X-X[initial]

## The Attempt at a Solution

I divided 7000N by 1500kg to cancel out the kg, to get 4.67 m/s(squared?) I assumed that was what my acceleration and said my final velocity should be 0. So my equation worked out to be 0=25+2×4.67(X-0). then I tried to solve for x and I got 2.67m, then I transposed 25 for acceleration and 4.67 for initial velocity and got .0467m which is way too small, the correct answer is 67m (according to masteringphysics). Even after getting the solution and plugging that into the equation for X ; the solution doesn't equal 0 like it should. I tried to solve using the position kinematics equation X=Xinitial+V(initial)×T(squared)+.5α×T(squared) but I couldn't make it work, since I wasn't given a time interval to work with.

## Answers and Replies

stockzahn
Homework Helper
Gold Member
Ahoi at PF!

It's really not easy to read mathematical work in words, but therefore symbols and signs can be used to make it more clear. So I try to translate what you've written at point 2):

## Homework Equations

I'm assuming it has something to do with F=mα and using that answer to solve for distance using the kinematics equation: Vfinal(squared)=Vinitial+2α[(X-X[initial]

##v_{final}^2 =v_{initial} +2as##

First of all, there is a small mistake in this formula - but I suppose it is a typo. Secondly, do you know where die formula comes from?

Merlin3189
Homework Helper
Gold Member
car mass =1500kg.
maximum friction force = 7000N.
I car is traveling at 25m/s,
what is the shortest distance the car can stop safely?

## Homework Equations

Vfinal(squared)=Vinitial+2α[(X-X[initial] Not a correct formula m2/sec2 ≠ m/sec + m2/sec2

## The Attempt at a Solution

I divided 7000N by 1500kg to cancel out the kg, to get 4.67 m/s(squared?)
So used F=ma transposed to a=F/m Ok to get 4.67 m/sec2 Ok
I assumed that was ...my acceleration and said my final velocity should be 0. Ok
So my equation worked out to be 0=25+2×4.67(X-0). but the formula is wrong, so this gives the wrong answer
then I tried to solve for x and I got 2.67m, ditto
then I transposed 25 for acceleration and 4.67 for initial velocity which is crazy! You don't just swap numbers around for no reason.
and got .0467m which is way too small, the correct answer is 67m (according to masteringphysics). I agree
Even after getting the solution and plugging that into the equation for X ; the solution doesn't equal 0 like it should. because that equation was wrong
I tried to solve using the position kinematics equation X=Xinitial+V(initial)×T(squared)+.5α×T(squared) but I couldn't make it work, since I wasn't given a time interval to work with. You were after the right formula originally, but just copied down wrongly.

I looked back at the kinematics formula and realized the initial velocity was also supposed to be squared (which I hadn't done, originally) and that's why my answer didn't match up.
I had transposed the numbers because I still didn't understand which calculation gave me the initial velocity and which gave me the acceleration, so I had hoped I had written them down wrong and what I had thought was my initial velocity was actually the acceleration and vice versa.

stockzahn
Homework Helper
Gold Member
I looked back at the kinematics formula and realized the initial velocity was also supposed to be squared (which I hadn't done, originally) and that's why my answer didn't match up.

That's good.

I had transposed the numbers because I still didn't understand which calculation gave me the initial velocity and which gave me the acceleration, so I had hoped I had written them down wrong and what I had thought was my initial velocity was actually the acceleration and vice versa.

There are some principles forming the basis for these calculations., one of them is energy conservation. In your case there are two forms of energy relevant: The kinetic energy of the car and the work done by the friction force. Since energy must be conserved, the two must be transformed into each other, so the difference of kinetic energy equals the work done by the friction force:

$$m_{car}\frac{v_{final}^2-v_{init}^2}{2} = F_{friction}s_{stop}$$

The friction force decelerates the car, that means its mass is negatively accelerated:

$$m_{car}\frac{v_{final}^2-v_{init}^2}{2} = m_{car}a_{deceleration}s_{stop}$$

Re-arrainging yields

$$v_{final}^2 = v_{init}^2+2a_{deceleration}s_{stop}$$

which is the formula you used. So it can be derived from the energy conservation principle, but what you need to use is its "original" form to solve the task.

I hope that helped.