Riding the Train -Vecotrs & components

[junesmrithi]

Riding the Train --Vecotrs & components

A passenger train is cruising directly to the north at 16.0 mi/hr.
what are the speed and (numerical) direction of a passenger with respect to the ground on this train, if he now walks east-ward at 5.4 mi/hr?
---Compass heading in degrees: N=0, E=90, S=180, and W=270

I know i have to add velocities add as vectors.

Work-
i drew a triagle, the y commponent being 16 and the x (hoizontal) component being 5.4 and i ahve to ifnd the resultant speed. aka: the hypothenuse of the triangle.
But i don't know how to do that when i don't have an angle.
Please help

 

[cristo]

Hint: Pythagoras' Theorem.

 

[m2003]

I would like to clarify a few things about vectors and components in this situation. Firstly, velocity is a vector quantity, which means it has both magnitude (speed) and direction. In this case, the train's velocity is 16.0 mi/hr due north, which can be represented as a vector pointing upwards on a graph.

Now, when the passenger starts walking eastward at 5.4 mi/hr, we can also represent this as a vector pointing to the right on the same graph. This is known as the passenger's velocity relative to the train.

To find the passenger's velocity with respect to the ground, we need to find the resultant vector by adding the two velocity vectors together. This can be done using vector addition, which involves finding the sum of the x-components and y-components separately.

In this case, the x-component of the train's velocity is 0 mi/hr (since it is moving directly north) and the x-component of the passenger's velocity is 5.4 mi/hr. So, the resultant x-component is 0 + 5.4 = 5.4 mi/hr.

Similarly, the y-component of the train's velocity is 16.0 mi/hr and the y-component of the passenger's velocity is 0 mi/hr (since the passenger is not moving vertically). So, the resultant y-component is 16.0 + 0 = 16.0 mi/hr.

Now, we can use the Pythagorean theorem to find the magnitude (or speed) of the resultant vector, which is the hypotenuse of the triangle you drew. The formula is c = √(a^2 + b^2), where c is the hypotenuse (resultant speed), a is the x-component (5.4 mi/hr), and b is the y-component (16.0 mi/hr).

Plugging in the values, we get:
c = √(5.4^2 + 16.0^2) = √(29.16 + 256) = √285.16 = 16.88 mi/hr

So, the passenger's speed with respect to the ground is 16.88 mi/hr. To find the direction, we can use trigonometric functions to find the angle between the resultant vector and the horizontal axis.

The tangent of this angle is the opposite side (16.0 mi/hr) divided by the adjacent side (