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pecosbill
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Homework Statement
A schoolbus travels around a circular path with acceleration a(t)=0.5t m/s/s with t in seconds.
At some point it has a velocity of 8 m/s.
What are the magnitudes of its velocity and acceleration when it has traveled a fourth of the circular track from the point at which it had v = 8 m/s?
Radius of the track is 250m
Homework Equations
dv=adt
ds=vdt
The Attempt at a Solution
Using the relation dv=adt,
dv=0.5tdt
I integrate both sides, but get confused as to where I should integrate from since the time the bus has a velocity of 8 m/s is not provided. After consideration, I choose:
v initial is 0, v final is 8 and t initial is 0, t final is just t
After integrating, I solve for t to find
t=6.32 at v=10 and v(t)=0.25t^2
then if a fourth of the circle is travelled, the distance traveled is a fourth of the circumference:
(2*pi*250)/4=125*pi
then ds=vdt=(0.03t^2)dt
We integrate again to find distance traveled as a function of t
integrate ds from 0 to 125*pi and vdt from 6.32 to t, yielding
125*pi=0.083(t)^3-0.083(6.32)^3
From this equation, t=17.08 seconds or it will take 17 seconds to travel a fourth of the way around from this point. From here, we can plug into the a(t) and v(t) equations to answer the question.
I think this is right; however, the first time I solved the problem, I integrated the dv=adt expression from 10 to v on the dv side and 0 to t on the adt side to get v(t)=0.25*t^2+10
I then integrated it again and found the distance as a function of time, solving the time it took to travel from the point at 10 m/s to a point a fourth of the way around the track. What is wrong with this approach?