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Physics 15 Online
OpenStudy (anonymous):

A subway train starting from rest leaves on with a constant acceleration. At the end of 9.34s it is moving at 15.0374 m/s. What is the train's displacement in the first 5.99628s of motion? Answer in units if m

OpenStudy (anonymous):

I think the Total time is,T=9.34 s Initial speed, u=0 Final speed, v = 15.0374 m/s

OpenStudy (anonymous):

Using fundamental motion equation v=u+aT a=(v-u)/T =(15.0374-0)/9.34 =1.61 m/s^2 Displacement at 5.99628 s s=ut+(1/2)at^2 =0+(1/2)*(1.61)*(5.99628)^2 =28.944 m

OpenStudy (anonymous):

Thanks so much for the help that's right :)

OpenStudy (anonymous):

your welcome :)

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