to find the first question, solve for when s(t) is zero. for the second, take the derivative of the first equation, and plug in the answer to the previous question
Calculus Question

to find the first question, solve for when s(t) is zero. for the second, take the derivative of the first equation, and plug in the answer to the previous question
He speaks the truth.

How would I solve this problem...
If a construction worker drops a wrench from a height of 1000 feet, when will the wrench hit the ground? At what velocity will the wrench impact the ground?
If I remember my physics:
The distance is 1/2 the acceleration (in this case, the acceleration of gravity) times the time squared. Or: 1000 feet = 0.5 X 32 feet/second squared X time squared. Solve for time. 1000/16 = t squared. T = 7.9 seconds.
Velocity is acceleration times time. v=at ; v = 32 x 7.9 ; v = (slightly greater than) 252 feet/second.
What does the calculus give you?

How would I solve this problem...
If a construction worker drops a wrench from a height of 1000 feet, when will the wrench hit the ground? At what velocity will the wrench impact the ground?
If I remember my physics:
The distance is 1/2 the acceleration (in this case, the acceleration of gravity) times the time squared. Or: 1000 feet = 0.5 X 32 feet/second squared X time squared. Solve for time. 1000/16 = t squared. T = 7.9 seconds.
Velocity is acceleration times time. v=at ; v = 32 x 7.9 ; v = (slightly greater than) 252 feet/second.
What does the calculus give you?
Same answer although I got -253 feet per second because the vector is facing down.

The calculus gives you a means to derive the formula for the wrench's velocity at a given time (t) armed with only the formula for it's position at a given time.
If you take the derivative yet again, you'll derive the formula for the wrench's acceleration at a given time (t), however you'll also find that it's a constant.

The calculus gives you a means to derive the formula for the wrench's velocity at a given time (t) armed with only the formula for it's position at a given time.
If you take the derivative yet again, you'll derive the formula for the wrench's acceleration at a given time (t), however you'll also find that it's a constant.
Well, that's probably what Newton did way back when. It's just easier for me to remember the formulas than the method to derive the formulas.

Yeah i know, like chess rocks alot!!
Chess is sicK, Its fun, addicting!! you name it!
Im so glad you made this thread in the "Chess General Discussion" because its all about chess!! WO0T!

I'll be honest, I'm impressed at how fast people are to help here.
Still, I think this website will be more help for the math/physics: (but definitely not for chess!) http://www.physicsforums.com/
I know it was a lifeline for me when I took Physics. I'd answer your limits question myself.... but it's been awhile since I've done limits.
Calculus fits in with chess because both calculus and chess are things done by smart people. I always wanted to do things that showed I was smart. I felt the lack of calculus so much that at age 50 I taught myself 1st year college calculus, going through all the problems in the book. If you ask me, playing a decent game of chess and knowing some calculus goes a long way towards showing that you are smart.
How would I solve this problem.
Use the position function s(t)=-16t^2+1000, which gives the height (in feet) of an object that has fallen for t seconds from a height of 1000 feet. The velocity at time t=a seconds is given by lim [s(a)-s(t)]/(a-t) as t>a.
If a construction worker drops a wrench from a height of 1000 feet, when will the wrench hit the ground? At what velocity will the wrench impact the ground?