position of a particle in the force field

Question

The force acting on a particle of unit mass is given in terms of time $t$ by $F = a \cos \omega t \hat\imath + b \sin \omega t \hat\jmath.$
If the particle is initially at rest at the origin, find its position at any later time.

I have tried to solve by following : by the second Newtons' law $m a = F.$ so $1 * a = F,$ where $$1* a = 1 * (d^2 r)/dt = F.$$ Integrating once we get the velocity : $$v = (a \sin \omega t)/ \omega \hat\imath - (b \cos \omega t) / \omega \hat\jmath$$ Integrating the second time we get the positon $$r = -a \cos (t \omega ) / \omega^2 \hat\imath - b \sin (t \omega ) / \omega^2 \hat\jmath$$

The answer in the book is different $$r = a (1 - \cos \omega t) / \omega^2 \hat\imath + b (\omega t - \sin \omega t) / \omega^2 \hat\jmath$$

Answer 1

Note that "initially at rest" means that $\mathbf{v}(0)=\mathbf{0}$. Hence $$\mathbf{v}(t)=\int_0^t (a \cos(ws) \mathbf{i} + b\sin (ws) \mathbf{j})ds=\frac{1}{w}\left[a \sin(ws) \mathbf{i} - b\cos (ws) \mathbf{j}\right]_0^t\\=\frac{a \sin(wt)}{w} \mathbf{i} - \frac{b(\cos (wt)-1)}{w} \mathbf{j}.$$ Can you take it from here?

Answer 2

$$\vec{v}(t)-\vec{v}(0)=\vec{v}(t)=\int_0^t (a\cos\omega t \hat{i}+b\sin\omega t \hat{j})dt=\frac{a}{\omega}\sin\omega t \hat{i}+\frac{b}{\omega}\left(1-\cos\omega t\right) \hat{j}$$ $$\vec{r}(t)-\vec{r}(0)=\vec{r}(t)=\int_0^t\left[\frac{a}{\omega}\sin\omega t \hat{i}+\frac{b}{\omega}\left(1-\cos\omega t\right) \hat{j}\right]dt$$ $$=\frac{a}{\omega^2}\left(1-\cos\omega t\right)\hat{i}+\left[\frac{b}{\omega}t-\frac{b}{\omega^2}\sin\omega t\right]\hat{j}$$