A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running directly east, what is Δx?

Guest May 9, 2019

#3**0 **

The y component of the acceleration is - .350 sin 52 = .=-.2758 m/s^2

When does the y component of the runners motion = 0?

v = v0 + at

0 = 2.88 + (-.2758) t so t = 10.44 sec

What is the speed in the x direction at this point in time?

x component of acceleration = .350 cos 52 = .2155 m/s^2

v in x direction =

v = vo + at

= 0 + .2155 (10.44) = 2.24 m/s

ElectricPavlov May 9, 2019