A passenger on a moving train walks at a speed of 2.50 m/s due north relative to

Question

A passenger on a moving train walks at a speed of 2.50 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 30.0 degree west of north. What are the magnitude and direction of the velocity of the train relative to the ground? X magnitude Think about the position vectors of the train (an arbitrary fixed point on it) and the passenger with respect to a fixed origin on the ground. What is the position vector of the passenger with respect to the train? How is it related to their positions with respect to the origin? Since velocity is position per unit time, what is the relationship between the relative velocity of the passenger with respect to the train and the velocities with respect to the origin? m/s direction degree west of north

Explanation / Answer

let north be vector j and east be vector i.
man's speed relative to ground=v1=4.5*(-sin 30 i + cos 30 j)

lets trains speed be v2 w.r.t. ground.

then relative velocity of man=v1-v2=2.5 j

so v2=v1-2.5 j

=-2.25 i +1.3971 j

magnitude=2.6485 m/s

direction=arctan(2.25/1.3971)=58.1625 degree west of north

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