beginners calculus application problem?

Question

alright so im real stuck on this one. somethings just not clicking with me. any help would be much appreciated. thanks!


Use the position function s(t) = -4.9t^2 + 150
which gives the height (in meters) of an object that has fallen from a height of 150 meters. the velocity at time t = a seconds is given by

lim [s(a) - s(t)] / [a - t]
t->a


Find the velocity of the object when t = 3


thanks again :)

Answer 1

Your limit formula is the definition of the derivative, but I suspect that you are supposed to solve it using limits, not derivative formulas. In that case, you just follow the equation, substituting the values of s(a) and s(t)

lim {[-4.9a^2 +150] - [-4.9t^2 + 150]}/(a - t)
t->a

Expanding the numerator -4.9a^2 + 4.9t^2 = 4.9*(t^2 - a^2) =

4.9*(t + a)*(t - a)

Then dividing by a - t you get -4.9(t + a); the limit as a->t of this is just -4.9*2*t = -9.8*3
When t = 3 the value becomes -3*9.96 or 29.4

Answer 2

s(t) = -4.9t^2 + 150
s'(t) = v = -9.8t

At t=3 v = -29.4

Another Method: Let t = 2.999 and a = 3.000

s(a) = -44.1 + 150 = 105.9
s(t) = 105.9293951

( s(a) - s(t) ) / (a-t)
-.0293951 / .001 = -29.3951

An answer closer to -29.4 will be obtained by making t closer to 3.000

Answer 3

velocity is the first derivative of position.
short cut here is 2 x -4.9m/s^2 x 3 sec = 29.4 m/s