How to find average acceleration? a bird is flying south at 8 m/s. the bird veers and changes velocity to 6 m/s east over a 2 second period. What is the average acceleration (magnitude and direction) during this 2 second interval?

Aug 23, 2016

$\left\mid \vec{a} \right\mid = 5 \frac{m}{s} ^ 2$

$\theta = \arctan \left(\frac{4}{3}\right)$

Explanation:

by definition average acceleration over time $t$ is:

$\vec{a} = \frac{\Delta \vec{v}}{t} = \frac{1}{t} \left({\vec{v}}_{2} - {\vec{v}}_{1}\right)$

using W-E as positive Cartesian x-direction and S-N as positive y direction, we have

${\vec{v}}_{1} = \left(\begin{matrix}0 \\ - 8\end{matrix}\right) , q \quad {\vec{v}}_{2} = \left(\begin{matrix}6 \\ 0\end{matrix}\right)$

$\implies \vec{a} = \frac{1}{2} \left(\left(\begin{matrix}6 \\ 0\end{matrix}\right) - \left(\begin{matrix}0 \\ - 8\end{matrix}\right)\right) = \frac{1}{2} \left(\begin{matrix}6 \\ 8\end{matrix}\right) = \left(\begin{matrix}3 \\ 4\end{matrix}\right)$

$\implies \vec{a} = \left(\begin{matrix}3 \\ 4\end{matrix}\right)$

from that we can calculate the magnitude of the average acceleration as

$\left\mid \vec{a} \right\mid = \sqrt{{3}^{3} + {4}^{2}} = 5 \frac{m}{s} ^ 2$

it's direction, where $\theta$ is measured positive anti clockwise from the x-axis is

$\tan \theta = \frac{4}{3} , \theta = \arctan \left(\frac{4}{3}\right)$