contestada

You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of reference he has a speed of 3 m/s to the south. What is the velocity of the man from the frame of reference of a stationary observer? is it 5m to the south??

Respuesta :

rameshafernandosxjtr

Yep the answer is 5ms^-1

Suppose that you as A and the other man as B,

Then, V(B,A) = V(B,E) - V(A,E)

          V(B,E) = V(B,A) + V(A,E)

                      = 3 + 2

                      = 5 ms^-1

Answer:

Other man speed is 5 m/s towards South

Explanation:

Our Velocity is given as 2 m/s towards South

So it is given as

[tex]v_1 = -2 \hat j[/tex]

from our frame the speed of other man is 3 m/s towards South

So it is given as

[tex]v_{21} = -3 \hat j[/tex]

now from the formula of relative velocity we can say

[tex]v_{21} = v_2 - v_1[/tex]

[tex]v_2 = v_1 + v_{21}[/tex]

[tex]v_2 = -2 \hat j - 3\hat j[/tex]

[tex]v_2 = - 5\hat j[/tex]

So other man velocity is 5 m/s towards South