A water balloon is launched upward with an initial velocity of $40 ft/s$. the height h at time $t$ is given $h(t)= 16t^2+40t$
a. find the avg. velocity of the balloon between $t=1$ and $t=2$ seconds
b. find the instantaneous velocity at $t=1$sec
I'm confused how to find the answers when given the formula for $h$..do you take the derivative and substitute $t$? If anyone can let me know how to set up each part of the problem that would be extremely helpful
$\endgroup$ 02 Answers
$\begingroup$a) $v_{ave} = \dfrac{h(2) - h(1)}{2 - 1} = \dfrac{16\cdot 2^2 + 40\cdot 2 - 16\cdot 1^2 - 40\cdot 1}{1} = 88 ft/s$.
b) $v_{int} = h'(1) = 32t + 40 |_{t = 1} = 32\cdot 1 + 40 = 72 ft/s$
$\endgroup$ $\begingroup$For the average velocity, draw a chord connecting the points where $t=1$ and $t=2$ on the graph. The average velocity will be slope of this chord.
To find the instantaneous velocity at $t = 1$, simply take the derivative and plug in $t=1$. Note that this will give you the slope of the tangent line to the graph at $t = 1$.
(This is the reasoning behind LAcarguy's response).
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