# The acceleration a(t), velocity v(t) and position s(t) of a particle moving along a coordinate line are given. If a(t)=e^t, v(0)=12 and s(0)=14 where is the partocle at time t

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### 2 Answers

a(t) = dv(t)/dt

a(t)dt = dv(t)

e^t dt = dv(t)

Integrate both sides will give you;

∫e^t dt = ∫dv(t)

e^t +C1 = v(t)-----(1)

It is given that v(0)=12

For (1) when t=0;

e^0+C1 = v(0)

1+C1 = 12

C1 = 11

Also v(t) = ds(t)/dt

Then v(t) dt = ds(t)

(e^t +11)dt = ds(t)

Integrate both sides will give you;

∫(e^t +11)dt = ∫ds(t)

e^t+11t+C2 = s(t)

It is given that s(0)=14

When time t=0;

Then e^0+11*0+C2 = s(0)

1+C2 = 14

C2 = 13

s(t) = e^t+11t+13

**At time t particle is at s(t) = e^t+11t+13**

The acceleration of the particle is `a(t) = (d^2s)/(dt^2) = e^t` .

`v(t) = (ds)/(dt) = int a(t) dt = int e^t dt = e^t + C`

`v(0) = 12`

=> `C = 11 `

`v(t) = e^t + 11`

`s(t) = int v(t) dt = int e^t + 11 dt = e^t + 11t + C2`

`s(0) = 14`

=> `C2 = 13`

The function `s(t) = e^t + 11t + 13`

**The position of the particle at time t is `s(t) = e^t + 11t + 13` **