What is the speed of a falling rock at t = 6 sec?

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Homework Help Overview

The problem involves determining the speed of a rock in free fall from a height of 400 meters, using the equation for motion on the planet Quixon, given as s = 3.8t². The inquiry centers around interpreting the equation and the relevance of the initial height in the context of the problem.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss whether the "m" in the equation is part of the equation or simply denotes meters, and how to properly apply the equation to find speed. There is also a debate about the relevance of the initial height of 400 meters and whether it is necessary for calculating speed at a specific time.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the interpretation of speed as instantaneous rather than average, but no consensus has been reached on the exact calculations or the role of the initial height.

Contextual Notes

Participants are navigating the implications of the problem's wording and the definitions of speed, as well as how to incorporate the initial height into their reasoning.

Jacobpm64
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The equation for free fall at the surface of the planet Quixon is s = 3.8t2m with t in sec. Assume a rock is dropped from the top of a 400-m cliff. Find the speed of the rock at t = 6 sec.
The problem is.. i don't know if the "m" is part of the equation or not.. like should i plug in the 400 as
s = 3.8t2(400) and then find the derivative and plug in 6 for t..
or should i just interpret the m as "meters" because s is a function to find position.. which would be in meters..
and in this case i'd just find the derivative of 3.8t2 and then plug in 6 for t...
hmm?
 
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Jacobpm64 said:
should i just interpret the m as "meters" because s is a function to find position.. which would be in meters..
and in this case i'd just find the derivative of 3.8t2 and then plug in 6 for t...
hmm?

I think that's what you should do.
 
then the 400 would be just extra information that is useless?
 
Jacobpm64 said:
then the 400 would be just extra information that is useless?

Well no it would be the initial height of the projectile, so if you were asked the height of the projectile after a time t it would be

400 - 3.8t^2
 
all right, but i wasn't asked the height :P.. but is the answer 45.6 m/s?
 
Jacobpm64 said:
all right, but i wasn't asked the height :P.. but is the answer 45.6 m/s?

Depending on their definition of speed it's either that or half that. If they mean speed as in distance traveled divided by time traveled then its 22.8, but if they mean speed as in absolute value of the instantaneous velociy then you're right.
 
Since they say "Find the speed of the rock at t = 6 sec." it should be clear that they are asking for instantaneous speed, not average speed.
 

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