How far does the car travel while slowing down?

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A 1000 kg car decelerates from 30.0 m/s to 23.4 m/s under a net force of 9500 N. To determine the distance traveled while slowing down, the work-energy theorem can be applied, where the work done equals the change in kinetic energy. The equation used is Fd = 1/2mvf^2 - 1/2mvi^2, with the initial and final velocities correctly identified as 30.0 m/s and 23.4 m/s, respectively. The negative work done indicates the car is slowing down, leading to a calculated distance of approximately 18.5 to 19 meters, considering significant figures. Proper attention to significant figures is advised, using the least number from the given data.
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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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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?
 
StephenDoty said:
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

?
 
StephenDoty said:
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

?
 
  • #10
yep.
 
  • #11
d= 18.5m or 19m with sig figs.?
 
  • #12
A general rule of thumb is to use the least amount of sig figs as the values given to you in the question.
 
  • #13
which would be 2 from 9500N
 
  • #14
do i have the right number of sig figs
 
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