Answer:
A ) displacement=75.69km
B) Angle[tex]= 28.92^o[/tex]
Explanation:
This is a trigonometric problem:
in order to answer A and B , we first need to know the total displacement on east and north.
the tricky part is when the car goes to the direction northeast, but we know that:
[tex]sin(\alpha )=\frac{opposite}{hypotenuse} \\where:\\opposite=north side\\hypotenuse=distance[/tex]
North'=9.86km
and we also know:
[tex]cos(\alpha )=\frac{adjacent}{hypotenuse} \\where:\\adjacent=east side\\hypotenuse=distance[/tex]
East'=18.54km
So know we have to total displacement
North=28km+North'=37.86km
East=50km+East'=68.54km
To calculate the total displacement, we have to find the hypotenuse, that is:
[tex]Td=\sqrt{North^2+East^2} =75.69km[/tex]
we can find the angle with:
[tex]\alpha = arctg (\frac {North} {East}) \\\\ \alpha = arctg (\frac {37.86} {68.54})= 28.92^o[/tex]
