Question

Some Physics questions! Please help!

Answer 1

1) 260 km/h

Let's use the following convention:

positive x-direction = east

positive y-direction = north

Here we have to find the north component of the velocity's airplane, which means we have to find its y-component.

We can use the formula:

$v_y = v sin \theta$

where

v = 750 km/h is the magnitude of the plane's velocity

$\theta=20.0^{\circ}$ is the angle between the direction of the plane and the positive x-axis

Substituting,

$v_y = (750)(sin 20)=256.5 km/h \sim 260 km/h$

2) $24^{\circ}$ north of east

In order to find the direction of flight, we have to consider that the vector representing the displacement of the plane is the hypothenuse of a right triangle, of which the displacements along the east and north direction are the sides.

Therefore, we have

$v_x = 220 km$ is the displacement towards east

$v_y = 100 km$ is the displacement towards north

Therefore, the angle that gives the direction is given by

$tan \theta = \frac{v_y}{v_x}$

And substituting,

$\theta =tan^{-1}( \frac{100}{200})=24^{\circ}$

and this angle is measured north of east.