problem is A train traveling a straight track at 2m/s, begins too accelerate a=(60v^-4)m/s^2, where v is in m/s. Determine its velocity and position 3s after the acceleration.
This is what I have.
a=dv/dt, int(dv/a)=int(dt)
integral from 2(m/s) to v of (dv/60v^-4) = integral of tdt from 0 to t.
then t=60[(v^5)/5] evaluated from 2 to v.
finally t = 12(v^5 - 32).
Solving for v, v=(t/12 +32)^(1/5) and substituting for t=3, v=2.003 m/s. (I am not sure about the V @ 3s, value seems to small).
To obtain the positions @ 3s:
v=ds/dt intds=intvdt
int from 0 to s of ds = int from 0 to t of (32+t/12)^(1/5)dt
s=[5/(1/12)]{32+t/12}^1/5 evaluated from 0 to t and @ t=3,
s=0.186
I think the values are too small so I wil appreciate if somebody can check the solutions.
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This is what I have.
a=dv/dt, int(dv/a)=int(dt)
integral from 2(m/s) to v of (dv/60v^-4) = integral of tdt from 0 to t.
then t=60[(v^5)/5] evaluated from 2 to v.
finally t = 12(v^5 - 32).
Solving for v, v=(t/12 +32)^(1/5) and substituting for t=3, v=2.003 m/s. (I am not sure about the V @ 3s, value seems to small).
To obtain the positions @ 3s:
v=ds/dt intds=intvdt
int from 0 to s of ds = int from 0 to t of (32+t/12)^(1/5)dt
s=[5/(1/12)]{32+t/12}^1/5 evaluated from 0 to t and @ t=3,
s=0.186
I think the values are too small so I wil appreciate if somebody can check the solutions.
[/img][/code]