contestada

A motorcycle has a velocity of 15 m/s, due south as it passes a carwith a velocity of 24 m/s, due north. What is the magnitude anddirection of the velocity of the motorcycle as seen by the driverof the car?

a. 9 m/s, north
b. 9 m/s, south
c. 15 m/s, north
d. 39 m/s, north
e. 39 m/s, south

Respuesta :

onyebuchinnaji

Answer:

(b) 9m/s south

Explanation:

Case 1: A motorcycle has a velocity of 15 m/s, due south

Case 2: A car with a velocity of 24 m/s, due north.

Let the velocity of the car due south = Vs↓

Let the velocity of the car due north = Vn↑

Magnitude of the velocity of the motorcycle as seen by the driver of the car = V

V = Vn - Vs

  = 24m/s - 15m/s = 9m/s↓

The magnitude and velocity of of the motorcycle as seen by the driver of the car = 9m/s south

The correct option is B

stigawithfun

Answer:

e. 39 m/s, south

Explanation:

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 relative to the car be = [tex]V_{MC}[/tex]

According to relativity of velocities in one dimension;

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

Now, take;

south to be negative (-ve)                  

north to be positive (+ve)

Therefore, we can say that;

[tex]V_{M}[/tex] = -15m/s        [since the velocity is due south]

[tex]V_{C}[/tex] = +24m/s       [since the velocity is due north]

Now, substitute the values of [tex]V_{M}[/tex] and [tex]V_{C}[/tex] into equation (i) as follows;

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

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

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

Remember that we have taken;

south to be negative (-ve)                  

north to be positive (+ve)

Since the result we got is negative, it means the speed is due south.

Therefore, the speed of the motorcycle as seen by the driver of the car is 3.9m/s, due south.

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