FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Posted (13 years 10 months 1 day 15 hours ago) and read 977 times:

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
(t -> a)

s(a) - s(t)
a - t

If anyone can explain to me how to do this I would be very appreciated of.
Thank you.

Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.

Jessman From United States of America, joined Jul 2001, 1506 posts, RR: 7
Reply 1, posted (13 years 10 months 1 day 14 hours ago) and read 957 times:

It is the basic definition of velocity, which is change in position divided by change in time.
If you travel 32 feet in one second your average velocity is 32ft/sec.
so s(a)-s(t) gives the change in position
and a-t gives the change in time.
I think your math teacher is trying to give you a concept with the lim(t->a) idea. This gives you the "derivitive" of the function. The derivitive of a function will give you it's instantaneous rate of change at a point. When you get into higher math you will find that the Position function is the starting point; the velocity function is the derivitive of the position function and the acceleration function is the derivitive of the velocity function. What's really fun is starting with acceleration and "integrating" to find position.
anyway the derivitive of -16t^2 - 1000 is -32t dt, that is the power of t multiplied by the coefficient of t and lowering the power of t by one and then the "dt" is simply change in t. You can disregard it for all calculation purposes.

Jessman From United States of America, joined Jul 2001, 1506 posts, RR: 7
Reply 2, posted (13 years 10 months 1 day 14 hours ago) and read 945 times:

Also, please excuse me It has been a while since I actually proved a derivitive. There are certain rules that I have just learned through Precalculus, Calculus I and Calculus II. If I remember correctly you could just plug in some numbers for a and t and make them increasingly closer together to prove that the fraction is true. Try t=1. Also shouldn't it be -16t^2 + 1000 because at t=0 that gives us the 1000 ft.

FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Reply 3, posted (13 years 10 months 1 day 14 hours ago) and read 937 times:

Ok. Now say I need to know the velocity after an object falls from 1000 ft for 5 seconds.

With the rate of change at 32, I can multiply that by 5 and get the answer, but that doesn't use the above equation.

If I use the above, that gets me 400ft (change in distance) over 5 seconds (change in t), which equals only 80, but the answer is 160. What am I doing wrong?

Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.

Jessman From United States of America, joined Jul 2001, 1506 posts, RR: 7
Reply 4, posted (13 years 10 months 1 day 14 hours ago) and read 932 times:

You're not. You're getting right answers. The situation is that the object is accelerating at a constant 32 ft/sec/sec. Therefore the object travels further in the 5th second of travel than in the first second of travel. It is also traveling faster at t=5 than at t=1. This also means it's average velocity from t=1 to t=5 is going to be less than it's true velocity at t=5.
Derivitive=instantaneous rate of change at a point.
at t=5 seconds the object is traveling 160ft/sec.
it's position has changed by the position formula though. so at t=5 the position has changed 400 feet. The average velocity is 80ft/sec. Those are all true statements. think of it as at t=1 second the object was traveling at 32ft/sec, at t=2; 64 ft/sec, at t=3; 96ft/sec...etc. as you see it was speeding up, so it's velocity was constantly changing.

Jessman From United States of America, joined Jul 2001, 1506 posts, RR: 7
Reply 5, posted (13 years 10 months 1 day 13 hours ago) and read 921 times:

Now I'm up to proving this.
say at t=1 and a =1.001
s(a)=983.967984
s(t)=984
a-t=.001
so the velocity formula in your first post give us an answer of -32.016, when t=1. If you notice as a-t gets smaller and smaller as it approaches zero at t=1 the answer will be 32.
now let t=2 and a=2.001
s(a)=935.935984
s(t)=936
a-t=.001
the velocity equation gives -64.016 and as a->t the equation at t=2 will head toward 64, which is 32t.

Anyway on a different topic, just to prove that the velocity a t=that equation put it this way. Velocity = change in distance over change in time. As change in time goes to zero you will find that, well, darn it it just works.