Question

In: Physics

A passenger on a moving train walks at a speed of 1.60 m/s due north relative...

A passenger on a moving train walks at a speed of 1.60 m/s due north relative to the train. The passenger's speed with respect to the ground is 4.5 m/s at an angle of 35.0° west of north. What are the magnitude and direction of the velocity of the train relative to the ground? _______m/s direction ______ ° west of north

Solutions

Expert Solution

let the direction be :

velocity of passenger with respect to group:

Vpg = Vpt + Vtg

Vtg = Vpg - Vpt

Lets resolve the vector Vpg into components

angle from positive x axis, θ = 125.0

Vpgx = Vpg * cos θ

= 4.5 * cos(125.0)

= -2.5811 m/s

Vpgy = Vpg * sin θ

= 4.5 * sin(125.0)

= 3.6862 m/s

Lets resolve the vector Vpt into components

angle from positive x axis, θ = 90.0

Vptx = Vpt * cos θ

= 16.0 * cos(90.0)

= 0 m/s

Vpty = Vpt * sin θ

= 16.0 * sin(90.0)

= 16 m/s

we have:

Vpg = -2.5811i+ 3.6862j

Vpt = 0i+ 16j

Vtg = Vpg - Vpt

Vtg = (-2.5811i+ 3.6862j) - (0i+ 16j)

= (Vpgx - Vptx)i + (Vpgy - Vpty)j

= (-2.5811 - 0)i + (3.6862 - 16)j

= -2.5811i- 12.3138j

magnitude of Vtg = sqrt (-2.5811^2 + -12.3138^2)

magnitude of Vtg = 12.6 m/s

Since x component is negative and y component is also negative

It lies in the 3rd cordinate

angle = 180 + atan(|y/x|)

angle = 180 + atan(|-12.3138/-2.5811|)

angle = 191.84 degree from x axis

which is 191.84 - 90 = 101.84 west of north

12.6 m/s

101.84 degree


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