Question

To a driver going east in a car with a velocity 40kn/h a bus appears to move towards north with a velocity of 40km /h what is the actual velocity and directon of the bus

Answer 1

Answer:

$|V_b |= 40\sqrt{2} kmph$

Explanation:

$\bold{ V_b =velocity \: of \: bus}$

$\bold{ V_d =velocity \: of \: driver}$

$\bold{ V_{bd} =velocity \: of \: bus \: w.r.t \: driver}$

Given:

$V_d = 40 i$

$V_{bd} = 40 j$

$V_b = ??$

Now,

$V _b - V_d = V_{bd}$

$V_b = 40 i + 40 j$

$|V_b |= \sqrt{{40}^{2}+{40}^{2}}$

$|V_b |= \sqrt{2({40}^{2})}$

$|V_b |= \sqrt{2(1600)}$

$|V_b |= \sqrt{3200}$

$|V_b |= 40\sqrt{2} kmph$

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°.