I have a huge problem

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wildthing1211

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Sep 25, 2005
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I have a word problem I don't get. A skydiver jumps from an airplane at an alltitude of 12,000 ft. After 42 seconds, she reaches 4606 ft. and opens her parachute. What was her average velocity during her descent? If anyone knows, that would mean a lot
 
time = 42 seconds

distance = 12000 - 4606 ft

average velocity = distance / time

You cannot do any calculation for the part after the chute opens as you do not have any data for this part.
 
I think they want to know the descent during this time without the parachute but I got 12,000-4,608 divided by 42=176 feet per sec.

Thanks anyway.
 
You know the displacement (∆d) and the change in time (∆t).

v(average) = ∆d/∆t = (4606-12000)/42 ≈ -1.8(10^2)m/s

That's the answer if you're rounding to 2 significant figures. The negative sign means the object is going down since velocity is a vector.
 
your answer of 176 ft/s is correct

please note your answer is a positive value as the reference plane is downwards (ie the positive direction is down as shown by your velocity values)
 
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