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

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Solution

Resultant displacement of the car $\to S = 30^i + 40^j$ km
$∴$ Magnitude of the displacement $| \to S | = \sqrt{30^{2} + 40^{2}} = 50$ km
Also $t a n θ = \frac{40}{30}$ $⟹ θ = 53^{o}$ (North of east)

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

Resultant displacement of the car →S=30^i+40^j km∴ Magnitude of the displacement |→S|=√302+402=50 kmAlso tanθ=4030 ⟹θ=53o (North of east)

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

Resultant displacement of the car $\to S = 30^i + 40^j$ km
$∴$ Magnitude of the displacement $| \to S | = \sqrt{30^{2} + 40^{2}} = 50$ km
Also $t a n θ = \frac{40}{30}$ $⟹ θ = 53^{o}$ (North of east)