Terminal Velocity of a sky diver

AI Thread Summary
The discussion revolves around calculating the acceleration of a sky diver at a speed of 30.0 m/s, given a mass of 83.0 kg and a terminal speed of 46.0 m/s. The initial approach used a linear resistance model (R = -bv), but it was suggested that fluid resistance is typically proportional to the square of velocity (R = -bv^2). The user received an incorrect answer in their homework applet, indicating potential issues with assumptions about resistance or significant figures. Clarification on whether to use the linear or quadratic model is essential for accurate calculations. The conversation highlights the common challenges faced in physics problems involving drag forces.
mkwok
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Homework Statement



A sky diver of mass 83.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 46.0 m/s.
(a) What is the acceleration of the sky diver when her speed is 30.0 m/s?

Homework Equations


R=-bv
mg-bv=ma


The Attempt at a Solution



since R is the resistance force acting on the sky diver: R=mg
therefore I set:
83*9.8 = -b(46),
and solve for b, b = -17.683
then plug the same numbers back into mg-bv=ma to solve for the acceleration at 30m/s
(83)(9.8)-(17.683)(30)=(83)a
a = 3.41m/s^2


but this is not correct, can someone please tell me what i did wrong?
 
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It looks fine to me. Though you usually take fluid friction proportional to v^2, not v.
 
well, so how would I approach this question with v^2?
I know for sure that I got the answer incorrect because I typed it into my online homework applet, it came out to be wrong.
 
You'd do it exactly the same but use R=-bv^2 instead. But all of this depends on what you are supposed to assume. Did the problem ask you to take R=-bv? This is always a problem with these applets. It could be expecting a minus sign (since the acceleration is down), it could be expecting a different number of significant figures, who knows? But it you think you should take R=-bv, then you did it correctly.
 
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