Question

a particle moves in the xy plane with a constant acceleration given by a =-4.0jm/s^2. At t=0, its position and velocity are 10im and (-2.0i+8.0j) m/s, respectively. What is the distance from the origin to the particle at t=2.0s?

Answer 1

Answer:

distance = 24.7m

Explanation:

$\overrightarrow{r(t)}=\frac{1}{2}\overrightarrow{a}t^2+\overrightarrow{v(0)}t+\overrightarrow{r(0)}=\frac{1}{2}\left[\begin{array}{c}0&4\end{array}\right]t^2+\left[\begin{array}{c}-2&8\end{array}\right]t+\left[\begin{array}{c}10&0\end{array}\right]$

$\overrightarrow{r(2)}=2\left[\begin{array}{c}0&4\end{array}\right]+2\left[\begin{array}{c}-2&8\end{array}\right]+\left[\begin{array}{c}10&0\end{array}\right]=\left[\begin{array}{c}6&24\end{array}\right]$