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Instantaneous Acceleration?

  • Thread starter tjsuglia
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  • #1
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A child on a Ferris wheel is moving vertically upward at 6.20 m/s at one instant, and 0.150 s later is moving at 6.20 m/s at an angle of 41.5° above the horizontal. Estimate the child's instantaneous acceleration.
What is the magnitude?
What is the direction? (degrees below the horizon)

I tried using the equation
ax=V(cos)THETA - Vinitial/TIME
ay=V(sin)THETA-0/TIME
Then taking the squareroot of the two answers above squared, but this did not work. where am i going wrong?
 

Answers and Replies

  • #2
PhanthomJay
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Think centripetal acceleration. The child is moving in a circle at constant speed (no tangential acceleration or tangential acceleration components)..
 
  • #3
ehild
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A child on a Ferris wheel is moving vertically upward at 6.20 m/s at one instant, and 0.150 s later is moving at 6.20 m/s at an angle of 41.5° above the horizontal.

I tried using the equation
ax=V(cos)THETA - Vinitial/TIME
ay=V(sin)THETA-0/TIME
Then taking the squareroot of the two answers above squared, but this did not work. where am i going wrong?
The angle is given with respect to the horizontal direction, which you used as x direction. Initially, the child moves vertically upward, so there is a non-zero initial velocity component in the y direction and the initial x component of the velocity is zero.

ehild
 

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