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

`d = 3574.3 m`

Step-by-step 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 + (1)/(2)a_x t^2`

`x = (83.5)(14) + (1)/(2)(8.5)(14)^2`

`x = 2002 m`

Displacement along y direction is given as

`y = v_y t + (1)/(2)a_y t^2`

`y = (-15.5)(14) + (1)/(2)(-28)(14^2)`

`y = -2961 m`

so the magnitude of the displacement is given as

`d = √(x^2 + y^2)`

`d = √(2002^2 + 2961^2)`

`d = 3574.3 m`