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Q3. A car is initially heading east at 55.0 km h when it turns a corner and heads N 55° Eat

65.0 km h -1. Determine the change in the velocity of the car during this time.​

User Zbynek
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1 Answer

2 votes

Answer:

37.3 km/h at N 2.70° E

Step-by-step explanation:

Find the horizontal and vertical components of the initial velocity (u) and final velocity (v). Remember that N 55° E means 55° east of north.

uₓ = 55.0 km/h

uᵧ = 0 km/h

vₓ = 65.0 km/h × sin 55° = 53.24 km/h

vᵧ = 65.0 km/h × cos 55° = 37.28 km/h

Find the difference.

vₓ − uₓ = -1.76 km/h

vᵧ − uᵧ = 37.28 km/h

The magnitude is:

|v − u| = √((-1.76 km/h)² + (37.28 km/h)²)

|v − u| = 37.3 km/h

The direction is:

θ = tan⁻¹(37.28 / -1.76)

θ = 92.70°

θ = N 2.70° E

User Nimrodshn
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