Question

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 32.0° west of north. What are the magnitude and direction of the velocity of the train relative to the ground?
magnitude _______m/sdirection ___________° north of east

Answer 1

Answer:

velocity = 3.25[m/s], angle = 47° or 137° north of east.

Explanation:

In order to solve this problem, we must make a diagram of the movement of each of the speed vectors, the vectors will be the velocities of the passenger and the train. It is of great importance to establish that the passenger's velocity with respect to the ground is the sum of the passenger's velocity with respect to the train plus the train's velocity with respect to the ground.

We draw a vector with a magnitude of 1.6 to the North (velocity of the person), then we draw another vector of 4.5 with an angle of 32 degrees to the North.

This velocity vector drawn is the total velocity relative to the ground. In such a way that if we combine the head of the first vector with the head of the second vector, we will obtain the velocity of the train with respect to the ground as well as its direction with respect to the North.

We have a velocity of 3.25 [m/s] at an angle of 47° respect to the north of west.

In the attached image we can find the graphic solution.