Question

You go watch your friend as she runs at a cross country meet. The first time she passes you, your friend runs by going due north at a speed of 4.40 m/s. You move to a different place on the course and twelve minutes after you saw her the first time, your friend runs past you again. This time she is traveling at a speed of 3.80 m/s in the direction 30.0° W of S. What is the direction of her average acceleration between those same two times?

Answer 1

76 degree south of west.

Step-by-step explanation:

We shall represent velocities in vector form , considering east as x axes and west as Y axes.

V₁ = 4.4 j

V₂ = 3.8, 30 degree west of south

V₂ = - 3.8 sin 30 i - 3.8 cos 30 j

= - 1.9 i - 3.29 j

Change in velocity

= V₂ - V₁

= - 1.9 i - 3.29 j - 4.4 j

= - 1.9 i - 7.69 j

Acceleration

= change in velocity / time

(- 1.9 i - 7.69 j ) / 60 ms⁻² .

Direction of acceleration θ

Tan θ = 7.69 / 1.9 =4.047

θ = 76 degree south of west.