Question

An escaped convict runs 1.70 km due East of the prison. He then runs due North to a friend's house. If the magnitude of the convict's total displacement vector is 2.50 km, what is the direction of his total displacement vector with respect to due East?

Answer 1

Answer:

The right approach will be "47° north of east".

Explanation:

The given values are:

East of prison

= 1.70 km

Displacement vector

= 2.50 km

Now,

The direction will be:

⇒  $Cos \ \theta =\frac{1.7}{2.5}$

⇒            $=0.68$

⇒         $\theta=47.16^{\circ}$

i,e.,       $\theta = 47^{\circ}$ (north of east)