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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?

 May 9, 2019
 #3
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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

 May 9, 2019

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