Question

The height (in feet) of a rocket from ground level is given by the function f(t) = -16t2 + 160t. What is the instantaneous velocity of the rocket 3 seconds after it is launched?

32 feet/second

64 feet/second

12 feet/second

10 feet/second

8 feet/second

Answer 1

hint:
we have to compute the first derivative of s(t), namely:
\[\Large v\left( t \right) = \frac{{d\left( { - 16{t^2} + 160t} \right)}}{{dt}} = ...?\]

Answer 2

oops..f(t), not s(t)

Answer 3

ok

Answer 4

d is?

Answer 5

it is a symbol. I can rewrite that formula, as below:
\[\Large v\left( t \right) = \left( { - 16{t^2} + 160t} \right)' = ...?\]

Answer 6

a 336

Answer 7

no, it is:
\[\Large v\left( t \right) = \left( { - 16{t^2} + 160t} \right)' = - 32t + 160\]
since the first derivative of t^2 is 2t, wheras the first derivative of t, is 1

Answer 8

whereas*

Answer 9

I got a 64

Answer 10

here are more steps:
\[\Large \begin{gathered}
v\left( t \right) = \left( { - 16{t^2} + 160t} \right)' = - 16 \times 2t + 160 \times 1 = \hfill \\
= - 32t + 160 \hfill \\
\end{gathered} \]

Answer 11

the requested velocity, is:
\[\Large v\left( 3 \right) = - 32 \times 3 + 160 = 64\;feet/\sec \]

Answer 12

so, your answer is right!

Answer 13

preciate it

Answer 14

thanks!