In: Physics
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
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