Question

When operating on Excepted Track, what speed must not be exceeded?

Answer 1

The speed of the man with respect to a passenger sitting at rest in the train is 12 m/s, as the directions of motion of the man and the train are perpendicular to each other.

Step-by-step explanation:

When determining the speed of a man with respect to a passenger sitting at rest in the train, we are dealing with relative motion in physics. If the man is running on a straight road which is perpendicular to the train track, and away from the track at a speed of 12 m/s, while the train is moving parallel to the track at a speed of 30 m/s, the speed of the man relative to a stationary passenger in the train is just the same as the man's speed on the ground, which is 12 m/s, because the motions are perpendicular to each other and, thus, do not interact with each other.

To visualize this, consider the train's motion to have no influence on the man's speed as viewed from the train if we are only considering speeds along the direction the man is running. Therefore, for a passenger at rest in the train, the man will appear to be moving straight away from the track at the same speed he would be observed by a stationary observer on the ground, assuming that the person on the train is not moving in the direction that the man is running.

Answer 2

The relative speed of the man with respect to a passenger sitting at rest in the train is approximately 32.5 m/s, which is calculated using the Pythagorean theorem to combine the perpendicular velocities of the man and the train.

Step-by-step explanation:

When we talk about the speed of one object in relation to another, we are dealing with relative speed. In the scenario presented where a man runs perpendicular to the train track at a speed of 12 m/s and a train is moving at 30 m/s, we're asked to calculate the speed of the man relative to a passenger sitting at rest in the train.

To solve it, we need to consider the Pythagorean theorem because the man and the train are moving in perpendicular directions. Let's denote the speed of the man with respect to the train as vmt. Using vector addition, vmt is the hypotenuse of a right-angled triangle with sides 12 m/s and 30 m/s, so vmt = √(122 + 302) m/s. After performing the calculation, we find that vmt equals approximately 32.5 m/s.

This type of question is a classic example of relative motion in a physics course, where understanding the vector nature of speed and velocity is crucial.