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Physics 9 Online
OpenStudy (akohl):

@Abhisar . A car of mass 1000 kg experiences a net force of 9500n while decelerating from 30m/s to 15.9m/s How far does it travel while slowing down?

OpenStudy (abhisar):

F=ma ---> a =F/m Now use the Third equation of motion to calculate distance. Take a=-F/m

OpenStudy (akohl):

So a =-9.5 m/s^2

OpenStudy (akohl):

What's the third equation of motion

OpenStudy (abhisar):

\(\large\sf v^2=u^2+as\)

OpenStudy (abhisar):

Refer this if it's not showing up http://prntscr.com/44ln9u

OpenStudy (akohl):

We have two v and no u

OpenStudy (abhisar):

V=final velocity (15.9 in this case) U=initial velocity (30 in this case)

OpenStudy (akohl):

I got 68m but that's not an answer on the multiple choice

OpenStudy (abhisar):

temme the answer choices

OpenStudy (akohl):

31 34 41 37

OpenStudy (abhisar):

I think ur calculation is wrong. I am getting 34m

OpenStudy (abhisar):

Recheck ur calculation

OpenStudy (akohl):

Got 34 this time thanks

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