Question

A spy report about a suspected car reads as follows."The car moved 2 km towards east, made a perpendicular left turn, ran for 500 m , made a perpendicular right turn, ran for 4 km and stopped." find the displacement of the car.

Answer 1

2 km East          0.50 km North       4km East again

Final displacement using the theorem of Pythagoras:
   √(6² + 0.50²) = √36.25  km

Answer 2

Answer:

The displacement of the car is 6.02 km.

Explanation:

Displacement of car AD = 2i + 0.5 j + 4i = 6i + 0.5 j   (where i & j are vector)

Magnitude of AD = $(A C)^{2}+(D C)^{2}=(A D)^{2}$

So $(A D)^{2}=(6)^{2}+(0.5)^{2}$ = 36 + 0.25 = 36.25

So AD = 6.02 Km

Now tan θ = DC/AC= 0.5 / 6 = 1/12

So $\theta=\tan ^{-1}(1 / 12)$

Hence displacement of the car is 6.02 Km along the direction$\tan ^{-1}(1 / 12)$ with positive K axis.