• Support PF! Buy your school textbooks, materials and every day products Here!

How far does the car travel while slowing down?

265
0
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
 

Answers and Replies

PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,120
463
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.
 
265
0
since we are on the energy and momentum chapter is there any way to do it with KE and PE?
 
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,120
463
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
[tex]W_{total} = W_{net} = \Delta KE[/tex]. 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:
265
0
so...
Fd=1/2mvf^2 - 1/2mvi^2

?????
 
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,120
463
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.
 
265
0
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.
 
265
0
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.
 
265
0
d= 18.5m or 19m with sig figs.?
 
458
0
A general rule of thumb is to use the least amount of sig figs as the values given to you in the question.
 
265
0
which would be 2 from 9500N
 
265
0
do i have the right number of sig figs
 

Related Threads for: How far does the car travel while slowing down?

Replies
1
Views
8K
Replies
5
Views
9K
Replies
14
Views
408
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
1K
Replies
2
Views
806
  • Last Post
Replies
6
Views
3K
Replies
2
Views
1K
Top