# How far does the car travel while slowing down?

A 1000.0 kg car experiences a net force of 9500N while decelerating from 30.0m/s to 23.4m/s. How far does the car travel while slowing down?

I do not even know how to start to setup the problem. I know that W=Fd but I do not know how to apply this with the velocities.

Any help would be appreciated. Thank you.

Stephen

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PhanthomJay
Homework Helper
Gold Member
You can calculate the car's acceleration from the given data using Newton's laws, then use the standard kinematic equations of motion to calculate the distance.

since we are on the energy and momentum chapter is there any way to do it with KE and PE?

PhanthomJay
Homework Helper
Gold Member
since we are on the energy and momentum chapter is there any way to do it with KE and PE?
Oh, sure. Use the work energy theorem
$$W_{total} = W_{net} = \Delta KE$$. In general , you've got to be a bit careful when using this equation, because W_net includes work done by both conservative (like gravity) and non-conservative (friction, etc.) forces, but it this case, those forces are given as one net number, so you don't have to worry about PE change.

Last edited:
so...
Fd=1/2mvf^2 - 1/2mvi^2

?????

PhanthomJay
Homework Helper
Gold Member
so...
Fd=1/2mvf^2 - 1/2mvi^2

?????
yes, where the value you use for F is the given F_net. Check it out both ways using the energy method vs. Newton 2 and the kinematic equations.

Would vi be 23.4m/s
and vf= 30m/s
???

If it is the other way around then F*D= a negative number

any help would be appreciated.

vf is 23.4m/s since you're slowing down, right? so the change in kinetic you get should be a negative number, and keep in mind that the force of friction is always negative, so the work done is negative too.

so vi= 30.0m/s
vf=23.4m/s

and the force of 9500N is negative???

-9500N*d=(1/2)(1000kg)(23.4m/s)^2 - (1/2)(1000kg)(30m/s)^2

???

yep.

d= 18.5m or 19m with sig figs.?

A general rule of thumb is to use the least amount of sig figs as the values given to you in the question.

which would be 2 from 9500N

do i have the right number of sig figs