Question

Q.7.
To a driver going east in a car with a velocity of 40 kmh, a bus appears to move
towards north with a velocity of 40root3 kmh! What is the actual velocity and
direction of motion of the bus?
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Answer 1

The train is going north at a velocity of 40√3 km/h and the car in going in an easterly direction at velocity 40 km/h

 Adding 40√3 km/h north and 40 km/h east, together

R = √[(40√3)² + 40²] = √6400

= 80km/h

 Direction is at an angle A north of east where:

 tanA = (40√3/40) A

= tan⁻¹ (√3) = 60

4.8

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Answer 2

Answer:

Given: Velocity of the car = 40km/h.

         Velocity of the bus w.r.t. the

          car = 40√3km/h

Vb = √(Vc + Vbc)^2

    = √[(40)^2 + (40√3)^2]

    = √(1600 + 4800)

    = √(6400)

    = 80km/h.

Direction of the motion of the bus....

Tan ß = 40/40√3

         = 1/√3

Tan ß = 30°.