Find the position of the particle. a(t) = t2 − 9t + 6, s(0) = 0, s(1) = 20 s(t)= A particle is moving with the given data. Find the position of the particle. a(t) = t2 − 9t + 6, s(0) = 0, s(1) = 20 s(t)=_______________
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You need to remember that position of particle may be found if the speed is given such that:
`s = v*t`
The problem provides you the function of acceleration, hence you should remember that `a(t) = (dv)/(dt),` hence `a(t)dt = dv` . You need to integrate the equation `a(t)dt = dv` such that:
`int a(t) dt = int dv =gt int (t^2-9t+6)dt=v(t)`
`v(t) = int t^2 dt - int 9t dt + int 6dt`
`v(t) = t^3/3 - 9t^2/2 + 6t + c_1`
You need to use the relation `v(t) = (ds)/(dt)` such that:
`ds = v(t)dt =gt int ds = int v(t)` dt
`s(t) = int(t^3/3 - 9t^2/2 + 6t + c_1) dt`
`s(t) = t^4/12 - 9t^3/6 + 6t^2/2 + c_1*t + c_2`
`s(t) = t^4/12 - 3t^3/2 + 3t^2 + c_1*t + c_2`
The problem provides the informations s(0)=0 and s(1)=20, hence, substituting 0 and 1 for t in s(t) yields:
`c_2=0`
`1/12 - 3/2 + 3 + c_1 = 20`
`c_1 = 17 + 3/2 - 1/12`
`c_1 = (204 + 18 - 1)/12`
`c_1 = 221/12`
Hence, evaluating the function s(t) under given conditions yields `s(t) = t^4/12 - 3t^3/2 + 3t^2 + 221t/12.`
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acceleration is rate of change of velocity = a(t) = d2x/dt2
I shall be using 'S' for integral sign in this question.
velocity = v(t) = S(d2x/dt2).dt = S(t2-9t+6)dt
= t3/3-9t2/2+6t+j where j is the constant of integration
distance = s(t) =S(v(t)).dt = S(t3/3-9t2/2+6t+j)dt
= t4/12-9t3/6+6t2/2+j.t+k where k is the constant of integration
we know that for t=0, s(0)=0
substituting these values in s(t) we get k = 0
also for t=1, s(1) = 20
substituting the values of k and s(1) in s(t) we get:
4/12-9/6+6/2+j = 20
or j= 20-1/3+3/2-3 = 109/3
subtituting the values of j, k in s(t) we get:
s(t) = t4/12-9t3/6+6t2/2+109t/3 or
s(t) = t4/12-3t3/2+3t2+109t/3
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