Question

a runner moves 2.88 m/s north. she accelerates at 0.350 m/s2 at -52.0 angle. at the point where she is running directly east, what is Δx?

Answer 1

11.7

Step-by-step explanation:

Answer 2

11.7 m

Step-by-step explanation:

I assume north is the y direction and x is the east direction, so Δx refers to the displacement in the east direction.

First, find the time it takes for the velocity to change from directly north to directly east.

Given (in the y direction):

v₀ = 2.88 m/s

v = 0 m/s

a = 0.350 m/s² sin(-52.0°) = -0.276 m/s²

Find: t

v = at + v₀

(0 m/s) = (-0.276 m/s²) t + (2.88 m/s)

t = 10.4 s

Given (in the x direction):

v₀ = 0 m/s

a = 0.350 m/s² cos(-52.0°) = 0.215 m/s²

t = 10.4 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10.4 s) + ½ (0.215 m/s²) (10.4 s)²

Δx = 11.7 m