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Physics 19 Online
OpenStudy (anonymous):

A train starts from rest and accelerates uniformly until it has traveled 5.6km and acquired a coward velocity of 42m/s. The train then moves at a constant velocity of 42m/s for 420 seconds. The train slows down uniformly at 0.65 m/s^2 until it is brought to a halt. What is the acceleration during the first 5.6 km. what is the total displacement for this motion?

OpenStudy (lucaz):

to find the acceleration for the first 5.6km you can use the formula d=vit+(at^2)/2 d=distance vi=initial velocity t=time a=acceleration

OpenStudy (lucaz):

sorry, you don't the time..

OpenStudy (anonymous):

Lucaz thank you so much for the reply! I found the answer to the first part. I need help with the displacement!

OpenStudy (lucaz):

since the acceleration is constant, you can use the average velocity to find the displacement

OpenStudy (anonymous):

How do I find the deceleration distance?

OpenStudy (lucaz):

the train starts to slow down with an initial velocity of 42m/s, the final velocity is 0, a=dv/dt, from this equation you find the change in time. now you can use the formula d=vit+(at^2)/2

OpenStudy (anonymous):

Vit being initial velocity?

OpenStudy (lucaz):

vit=initial vel times change in time.. a=dv/dt -0.65=(vf-vi)/dt dt=(0-42)/-0.65=64.6 approx.

OpenStudy (lucaz):

d=vit+(at^2)/2 d=42(64.6)+[(-0.65)(64.6)^2]/2=1356.923m

OpenStudy (lucaz):

if you have a calculator check my answer.. I think thats it

OpenStudy (anonymous):

Still a little confused on this one. Would you mind explaining this a different way? More as to in words rather than equations. I'm so wrapped up in this problem I just can't figure it out!

OpenStudy (lucaz):

one sec.

OpenStudy (lucaz):

I don't know exactly how to explain without the equations.. you can use these to solve this problem:

OpenStudy (anonymous):

!!! I got the right answer, I was just wanting a clearer understanding conceptually of what I was doing, rather than plugging in numbers into equations.

OpenStudy (lucaz):

|dw:1410390649295:dw|

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