Question

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?

Answer 1

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 $V_{M}$

Let the velocity of the car be $V_{C}$

Let the velocity of the motorcycle as seen by the car be $V_{MC}$

Using the principle of relativity;

$V_{MC}$ = $V_{M}$ - $V_{C}$       -------------------------(i)

From the question;

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

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

Substitute these values into equation (i) as follows;

$V_{MC}$ = -24 - (+15)

$V_{MC}$ = -24 - 15

$V_{MC}$ = - 39 m/s

Since the result of $V_{MC}$ 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.