Question

A jet plane is flying at a constant altitude. At time t1=0t 1=0, it has components of velocity vx=90m/s,vy=110m/sv x = 90m/s,v y=110m/s. At time t2=30.0st 2=30.0s, the components are vx=−170m/s,vy=40m/sv x =−170m/s,v y=40m/s.
(a) Sketch the velocity vectors at t1and t2.
How do these two vectors differ? For this time interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.

Answer 1

The average acceleration $\vec a_{\rm ave}$ over some time interval $[t_1,t_2]$ is equal to the ratio of the change in velocity $\vec v_2-\vec v_1$ over the duration of the interval $t_2-t_1$, or

$\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}$

which can be split into the $x$ and $y$ components as

$a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}$

$a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}$

The magnitude of this average acceleration is

$\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}$

and its direction is $\theta$ such that

$\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ$

which corresponds to a direction of about 15.1º South of West.