Consider a particle moving along the x-axis where x(t) is the position of the particle at time t A particle, initially at rest, moves along the x-axis such that its acceleration at time t > 0 is given by a(t)=5cos(t). At t=0, its position is x=2?

`a=(dv)/dt and v=dx/dt`
so with `a(t)=5cost` we have:

`\ \ \ \ \ (dv)/dt = 5cost`
`:. v=5sint+A`

`v=0` when `t=0` (initially at rest); so

`\ \ \ \ \ 0 = 0+A => A=0`
`:. v=5sint`

So then:

`\ \ \ \ \ (dx)/dt = 5sint`
`:. x=-5cost+B`

`x=2` when `t=0`; so

`\ \ \ \ \ 2 = -5+B => B=7`
`:. x=-5cost+7`

If the particle is at rest then `v=0`

`:. 5sint = 0`
`:. sint = 0`
`:. t = npi` where `n in NN`

Note:
Some tutors and texts combine the initial conditions into a definite integral and remove the need to evaluate the constant of integration, as follows:
e.g for the velocity we have:

`(dv)/dt = 5cost => int (dv)/dt dt=int5costdt`

If we combine the initial conditions into a definite integral then the integration variable `t` is arbitrary, then we can write

`int_0^v (dv)/dt dt=int_0^t 5costdt`
`:. int_0^v dv=int_0^t 5costdt` (or if you prefer `int_0^v d rho=int_0^t 5costau d tau`)
`:. [v]_0^v= [5sint]_0^t`
`:. v-0= 5sint - 5sin0`
`:. v= 5sint`, as before

Similarly for the displacement we could write

`int_2^x dx= int_0^t 5sint`, as before

and again we get the same solution.