Question

A jet plane is flying at a constant altitude. At time t₁ = 0 it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t₂ = 30.0s the components are vx = -170m/s, vy = 40m/s.

(a) Sketch the velocity vectors at t and t₂ . How do these two vectors differ?

(b) For the above time interval in part (a) calculate the components of the average acceleration,

(c) and the magnitude and direction of the average acceleration.

Answer 1

Given:

$v_{x1}=90\:{\rm m/s},\quad v_{y1}=110\:\rm m/s$

$v_{x2}=-170\:{\rm m/s},\quad v_{y2}=40\:\rm m/s$

(a)

(b) the components of the average acceleration

$$a_x=\frac{v_{x2}-v_{x1}}{t_2-t_1}=\frac{(-170-90)\:\rm m/s}{30.0\:\rm s}=-8.67\:\rm m/s^2$$

$$a_y=\frac{v_{y2}-v_{y1}}{t_2-t_1}=\frac{(40-110)\:\rm m/s}{30.0\:\rm s}=-2.33\:\rm m/s^2$$

(c) the magnitude of the average acceleration

$$a=\sqrt{a_x^2+a_y^2}\!=\!\sqrt{(-8.67)^2\!+\!(-2.33)^2}\!=\!8.98\:\rm m/s$$

the direction of the average acceleration

$$\theta=\tan^{-1}\frac{a_y}{a_x}=\tan^{-1}\frac{-2.33}{-8.67}=195^{\circ}$$