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A motorcycle has a velocity of 24 m/s, due south as it passes a car with a velocity of 15 m/s, due north. What is the magnitude and direction of the velocity of the motorcycle as seen by the driver of the car?

Respuesta :

stigawithfun

Answer:

39m/s due south.

Explanation:

Let's take the direction due south to be negative(-ve) and

Let's take the direction due north to be positive (+ve)

Also,

Let the velocity of the motorcycle be [tex]V_{M}[/tex]

Let the velocity of the car be [tex]V_{C}[/tex]

Let the velocity of the motorcycle as seen by the car be [tex]V_{MC}[/tex]

Using the principle of relativity;

[tex]V_{MC}[/tex] = [tex]V_{M}[/tex] - [tex]V_{C}[/tex]       -------------------------(i)

From the question;

[tex]V_{M}[/tex] = 24m/s due south = -24m/s    [since the south direction is -ve]

[tex]V_{M}[/tex] = 15m/s due north = +15m/s    [since the north direction is +ve]

Substitute these values into equation (i) as follows;

[tex]V_{MC}[/tex] = -24 - (+15)

[tex]V_{MC}[/tex] = -24 - 15

[tex]V_{MC}[/tex] = - 39 m/s

Since the result of [tex]V_{MC}[/tex] is negative, that means its direction is due south.

Therefore, the velocity of the motorcycle as seen by the car is 39m/s due south.

 

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