Question

A bus X is travelling with a speed of 40 km/hr towards North.Another bus Y is travelling with a speed of 30km/hr towards East. The velocity of bus X with respect to bus Y is?

Answer 1

Bus Y will note bus X to be travelling at 50 km/hr in a direction 53.1° north of west.

Explanation:

This is a 2-dimensional problem, and must be done by a careful vector addition.

We start by considering what the motion of X would be relative to Y if bus X were stationary. To Y, bus X would seem to be travelling toward the west at 30 km/hr.

Next, we add the actual speed of bus X, namely 40 km/hr to the north.

These velocity vectors are perpendicular to each other, and so, to add them, we must use Pythagorus' theorem.

`v_t^2 = v_X^2+v_Y^2`

`v_t^2 = 30^2+40^2=2500`

So, `v_t=sqrt2500 = 50` km/hr.

To get the direction of this velocity, we use `tan^(-1)`:

`theta=tan^(-1)(40/30)=53.1°`

You must be careful in interpreting this direction, however, as the 53.1° angle here would be measured clockwise from the direction of west. We can call this W53.1°N or 53.1° north of west.
As a standard angle (counter-clockwise from east) the angle would be the supplement of this, namely 143.1°.