a particle moves along an s-axis, use the given information to find the position function of the particle. `a(t)=2t^-3` ; `v(1)=2` ; `s(1)=18`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`a(t) = 2t^(-3)`

Acceleration is the rate of change of velocity with time.

`a(t) = (dV(t))/(dt)`

`dV(t) = a(t)dt`

Using integration;

`intdV(t) = inta(t)dt`

V(t) = int2t^(-3)dt

`V(t) = 2xxt^(-2)/(-2)+C` where C is a constant.

`V(t) = -t^(-2)+C`

It is given that` V(1) = 2`

`V(t) = -t^(-2)+C`

`V(1) = -1^(-2)+C`

`2 = -1^(-2)+C`

`C = 3`

`V(t) = -t^(-2)+3`

Velocity is the rate of change of position.

`V(t) = (d(S(t))/(dt)`

`dS(t) = V(t)dt`

`S(t) = intV(t)dt`

`S(t) = int(-t^(-2)+3)dt`

`S(t) = -t^(-1)/(-1)+3t+D` where D is constant.

It is given that `S(1) = 18`

`18 = -1^(-1)/(-1)+3xx1+D`

`D = 14`

`S(t) = 1/t+3t+14 `

So the position function of the particle is `S(t) = 1/t+3t+14`

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