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Updated on: 01 Oct 2023, 12:58
Originally posted by mikemcgarry on 08 Oct 2016, 15:16.
Last edited by BottomJee on 01 Oct 2023, 12:58, edited 3 times in total.
Last edited by BottomJee on 01 Oct 2023, 12:58, edited 3 times in total.
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D: 12
v: 30
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (03:10) correct
54%
(02:48)
wrong
based on 104
sessions
History
Date
Time
Result
Not Attempted Yet
On planet K, there is essentially no atmosphere, so objects in freefall experience no air friction. At time t = 0, an object released from rest accelerates downward at a uniform rate: its downward speed uniformly increases. At time t = 2 second, the object has fallen a distance D in meters from its original height. At time t = 5 second, the object is 75 meters below its original height, and is moving at a downward speed of v in meters per second.
In the table below, select values of D, the distance below the initial height at t = 2 seconds, and v, the downward speed at t = 5 seconds, that are consistent with the information provided. Make only two selections, one in each column.
In the table below, select values of D, the distance below the initial height at t = 2 seconds, and v, the downward speed at t = 5 seconds, that are consistent with the information provided. Make only two selections, one in each column.
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| D | v | |
| 12 | ||
| 30 | ||
| 60 | ||
| 75 | ||
| 90 | ||
| 150 |
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Official Answer
D: 12
v: 30
14 Oct 2016, 15:10
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Hi mikemcgarry,
t=5 distance is 75
75/5 = 15 meters per second.
But we don't have that in option. is it because "accelerates downward at a uniform rate" ?
t=5 distance is 75
75/5 = 15 meters per second.
But we don't have that in option. is it because "accelerates downward at a uniform rate" ?
14 Oct 2016, 15:58
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Shrivathsan
Dear Shrivathsan,
I'm happy to respond.
When you divide total distance by total time, that's average velocity over the entire five second interval. You have 100% correctly calculated the average velocity, but unfortunately, that's not the "v" for which the question was asking. You perfectly calculated the answer for a question that was not asked.
Does this make sense?
Mike
I used physics to solve this. 
