Answer to Question #215790 in Physics for Aru

Question

A jet plane is flying at a certain altitude. At time t1 = 0, it has components of velocity Vx = 90 m/s, Vy = 110 m/s. At time t2 = 30 sec, the components are Vx = -170 m/s, Vy = 40 m/s. Calculate (a) the components of average acceleration, (b) the magnitude and direction of average acceleration.

Answer 1

By definition, the average acceleration vector is given as follows:

$$\mathbf{a} = \left(\dfrac{v_{xf} - v_{xi}}{t_2 - t_1 } , \dfrac{v_{yf} - v_{yi}}{t_2 - t_1 }\right)$$

where $v_{xi} = 90m/s, v_{xf} = -170m/s$ are initial and final x-compenents of the velocity respectively, $v_{yi} = 110m/s, v_{yf} = 40m/s$ are initial and final y-compenents of the velocity respectively, $t_1 = 0, t_2 = 30s$.

Thus, obtain the following components:

$$a_x = \dfrac{v_{xf} - v_{xi}}{t_2 - t_1 } = \dfrac{-170-90}{30 -0 } \approx -8.67m/s^2\\ a_y = \dfrac{v_{yf} - v_{yi}}{t_2 - t_1 } = \dfrac{40- 110}{30 -0 } \approx -2.33m/s^2$$

The magnitue is given as follows:

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

And the direction (angle with the positive x-axis):

$$\theta = \arctan\left( \dfrac{a_y}{a_x} \right) + 180\degree = \arctan\left( \dfrac{-2.33}{-8.67} \right) +180\degree \approx 195\degree$$

Answer. a) $a_x = -8.67m/s^2, \space a_y = -2.33m/s^2$. b) $a = 8.98m/s^2,\space \theta = 195\degree.$