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rate of change question help plssss

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RuiAce:

--- Quote from: aadharmg on July 14, 2017, 04:43:07 pm ---Thanks for the guidance, I was able to make an equation for x in terms of t. The problem now is that because t is in seconds, when x = 0, t must be = 120 because after 2 minutes it reaches the ground. To find the constant after integrating the velocity equation, when I substitute 120 for t and 0 for x, I get a really weird number. If I carry that number, when I substitute t = 0 to find H, I end up with a negative and bizarre answer.

--- End quote ---
I think I realised the problem.

What usually happens in mechanics is that the orientation flips when you're considering upwards or downwards motion. I think in this scenario, you consider downwards as the positive motion.

So instead what you should be doing is treating the highest you can be (the cloud) the origin, i.e. x=0. So when t=0, x=0. Then when t=120, x=H, i.e. H is how far down you are from the cloud.

(Pardon the late reply - still doing stuff for lectures.)

aadharmg:

--- Quote from: RuiSmash on July 14, 2017, 05:27:04 pm ---I think realised the problem.

What usually happens in mechanics is that the orientation flips when you're considering upwards or downwards motion. I think in this scenario, you consider downwards as the positive motion.

So instead what you should be doing is treating the highest you can be (the cloud) the origin, i.e. x=0. So when t=0, x=0. Then when t=120, x=H, i.e. H is how far down you are from the cloud.

(Pardon the late reply - still doing stuff for lectures.)

--- End quote ---

Yessssss, I also finally understand. Thank you so much for the help, considering I have trials coming up, it's great that I can now tackle such a problem, or at least understand it. Thankkkkssssss! (PS the late replies weren't noticed at all, so it's all good)

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