A mobile moves horizontally on the $x$-axis (graduated in meters). Its speed as a function of time is given by $v(t) = 6 e^{\frac{-2t}{5}} \cos 5t$ and its initial position is $s (0) = 3$. Determine its final position and the distance it will have travelled.
Could anyone be able to complete this exercise? I'd like to confirm what I've done so far.
The position is $s(t)=\frac{-30 e^{\frac{-2t}{5}} (2 \cos 5t - 25 \sin 5t)+1947}{629}$.
Do I have to compute the time when the speed is $0$ for the final position and use the arc length ($\int_a^b \sqrt{1+f(t)^2} dt$) to compute the distance travelled? The answers I've obtained are $3.09214$ for the position and $3.18954$ for the distance travelled.
We may define distance travelled $s(t)$ as
$$s(t) =\int_{t_i}^{t_f}|v(t)|dt$$
So you have to split the integral wherever $v(t)=0$
If you have to find the total distance travelled starting from $t = 0$ to $t \to \infty$,
$$S=\int_{0}^{\infty}|v(t)| dt$$
I computed $S$ on Wolfram Alpha and it came out to be close to 9.55:
For finding final position, you may compute:
$$\lim_{t\to\infty}\frac{-30 e^{\frac{-2t}{5}} (2 \cos 5t - 25 \sin 5t)+1947}{629}$$
which comes out to be $\frac{1947}{629}$.