A car travels due east on a level road for `30 km`. It then turns due north at an intersection and travels `40 km` before stopping. Find the resultant displacement of the car.

Options

A. ` 50 km, 53^(@) ` north of east
B. ` 50 km, 53^(@) ` east of north
C. ` 100 km, 37^(@) ` north of east
D. ` 100 km, 37^(@) ` east of north

Answer

A

Explanation

Vector `vecA` represents a displacement of 30 km due east and vector `vecB` represents a displacement of 40 km due north. Therefore resultant displacement of the car is `vecC`

Now its magnitude is
`|vecC|=sqrt(A^2+B^2)=sqrt((30)^2+(40)^2)=50 km `
The direction of the resultant displacement is given by
`tan theta = B / A = (40)/30 = 1.33 " or " theta = tan^(-1) (1.33) = 53^@`
Thus the resultant displacement has a magnitude of 50 km and makes an angle `53^@` north of east .