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Old August 26 2009, 06:44 AM   #1
MIB
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I need some calculus help.

After spending a good hour and a half on some homework there is still one lingering question that keeps thwarting my attempts to answer. (Or, more accurately, my answers aren't being accepted by the math study program I'm using.) If anyone can point me in the right direction with this one, I'd greatly appreciate it. The problem reads as follows:

Use the position function s(t) = –16t^2 + 750, which gives the height (in feet) of an object that has fallen for t seconds from a height of 750 feet. The velocity at time t = a seconds is given by the following.

Limit as t -> a
(S(a) - S(t)) / (a - t)

If a construction worker drops a wrench from a height of 750 feet, how fast will the wrench be falling after 3 seconds?
The hell of it is that the question after this one is very similar and I answered it just fine without resorting to asking for help.
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Old August 26 2009, 07:25 AM   #2
Iasius
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Re: I need some calculus help.

I'm not really sure what I can write here that isn't pretty much the answer already.

Velocity is the first derivative of position is what ...
The velocity at time t = a seconds is given by the following.

Limit as t -> a
(S(a) - S(t)) / (a - t)
... is saying.


Point being, I don't really understand your problem with this if you don't explain a bit.
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Old August 26 2009, 07:49 AM   #3
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Re: I need some calculus help.

negative (nine times 16) 16X9=acceleration going to say 16X3=velocity just because LOL you need velocity right feet per sec not feet per sec squared

? Don't make me get the physics book for some formula I have intensionally forgotten.
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Old August 26 2009, 07:53 AM   #4
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Re: I need some calculus help.

It may be easier to understand if you try plugging in values for a and t that are both very close to three.
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Old August 26 2009, 08:06 AM   #5
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Re: I need some calculus help.

Don't they teach equations such as "v = u + a.t" and the Metric system any more?

Anyway, plugging in the values gives s(3) = 606 ft, and so (606 - 750)/(3 - 0) = -48 ft/s is your answer. Just that v = 0 - 16.3 = -48 ft/s seems a lot more direct way of getting to it.

Might not stop your Polar Explorer crashing into Mars, 'though.
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Old August 26 2009, 08:14 AM   #6
Iasius
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Re: I need some calculus help.

Great Mambo Chicken wrote: View Post
Don't they teach equations such as "v = u + a.t" and the Metric system any more?

Anyway, plugging in the values gives s(3) = 606 ft, and so (606 - 750)/(3 - 0) = -48 ft/s is your answer. Just that v = 0 - 16.3 = -48 ft/s seems a lot more direct way of getting to it.

Might not stop your Polar Explorer crashing into Mars, 'though.
That's the average speed over the first three seconds, not the velocity after three seconds.

v(t) = ds(t) / dt = -32t

The velocity after three seconds is v(3) = -96.

Don't they teach s(t) = 1/2 * a * t² + v0 * t + h anymore?
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Old August 26 2009, 08:37 AM   #7
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Re: I need some calculus help.

Lol. I won't be sending any probes to Mars.
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Old August 26 2009, 03:22 PM   #8
MIB
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Re: I need some calculus help.

I appreciate the help, guys. However, I was actually able to solve this one afterall. Which, in turn, underlined how important sleep is. I sat and stared at this problem for 30 minutes while I was dead tired (big mistake). Finally, I went to bed and after getting some sleep I was able to solve it with no problem within 30 seconds.

Again though, I do appreciate your help. The Mars probes no doubt appreciate it as well.
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Old August 26 2009, 04:01 PM   #9
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Re: I need some calculus help.

MIB wrote: View Post
I appreciate the help, guys. However, I was actually able to solve this one afterall. Which, in turn, underlined how important sleep is. I sat and stared at this problem for 30 minutes while I was dead tired (big mistake). Finally, I went to bed and after getting some sleep I was able to solve it with no problem within 30 seconds.

Again though, I do appreciate your help. The Mars probes no doubt appreciate it as well.
This is good advice for life in general. If something's stumping you, walk away from it for a while. Get something to eat. Get some rest. Do something fun. Do anything but think directly about the problem. You'd be surprised how often something just pops into your head and you realize you know what to do now.

I have a whole list of "oblique strategies" for this very purpose.
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Old August 26 2009, 06:55 PM   #10
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Re: I need some calculus help.

Robert Maxwell wrote: View Post
This is good advice for life in general. If something's stumping you, walk away from it for a while. Get something to eat. Get some rest. Do something fun. Do anything but think directly about the problem. You'd be surprised how often something just pops into your head and you realize you know what to do now.
Indeed! I concur wholeheartedly! -very good advice
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Old August 26 2009, 07:58 PM   #11
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Re: I need some calculus help.

Iasius wrote: View Post
Great Mambo Chicken wrote: View Post
Don't they teach equations such as "v = u + a.t" and the Metric system any more?

Anyway, plugging in the values gives s(3) = 606 ft, and so (606 - 750)/(3 - 0) = -48 ft/s is your answer. Just that v = 0 - 16.3 = -48 ft/s seems a lot more direct way of getting to it.

Might not stop your Polar Explorer crashing into Mars, 'though.
That's the average speed over the first three seconds, not the velocity after three seconds.

v(t) = ds(t) / dt = -32t

The velocity after three seconds is v(3) = -96.

Don't they teach s(t) = 1/2 * a * t² + v0 * t + h anymore?
I agree with this solution.

s(t) = 1/2*at^2 + V0*t + h. In other words s(t)=h.

Velocity is the rate at which the position is changing. So, take the derivative with respect to time.

v(t) = ds/dt = at = -32t

Acceleration is the rate at which the velocity is changing. So if you ever have to find that, take the derivative of v(t).

a(t) = d^2s/dt^2 = dv/dt = a = -32 ft/s^2, which is the actual rate of acceleration at the Earth's surface.
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