Question

An airplane in level flight travels horizontally with a constant eastward acceleration of 8.50 m/s2 and a constant northward acceleration of −28.0 m/s2. The airplane's initial velocity has eastward and northward components of 83.5 m/s and −15.5 m/s, respectively. Determine the magnitude of the airplane's displacement from its initial position after 14.0 s.

Answer 1

Answer:

$d = 3574.3 m$

Explanation:

Given that acceleration of the airplane is

$a_x = 8.50 m/s^2$

$a_y = -28 m/s^2$

initial velocity is given as

$v_x = 83.5 m/s$

$v_y = -15.5 m/s$

now we have displacement in x direction given as

$x = v_x t + \frac{1}{2}a_x t^2$

$x = (83.5)(14) + \frac{1}{2}(8.5)(14)^2$

$x = 2002 m$

Displacement along y direction is given as

$y = v_y t + \frac{1}{2}a_y t^2$

$y = (-15.5)(14) + \frac{1}{2}(-28)(14^2)$

$y = -2961 m$

so the magnitude of the displacement is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{2002^2 + 2961^2}$

$d = 3574.3 m$