Question

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Answer 1

take the anti derivative twice, starting with \(a(t)=30\) so \(v(t)=30t+C\)

Answer 2

since initial velocity is \(-10\) i.e. \(v(0)=-10\) you know \(v(t)=30t-10\] then repeat

Answer 3

oops i meant \[v(t)=30t-10\] then take the anti derivative again

Answer 4

so 15t^2 -10t+c

Answer 5

?

Answer 6

yeah but you know C too

Answer 7

since the "initial position" is 4, that means \[s(t)=15t^2-10t+4\]

Answer 8

initial position is what you get when \(t=0\) in other words it is the constant

Answer 9

so s(t)= 4?

Answer 10

@satellite73

Answer 11

4 ft/sec?

Answer 12

..i dont think thats right but im not sure

Answer 13

you know that
\(s(t)=15t^2-10t+C\)

and that \(s(0)=15(0)^2-10(0)+C \color{blue}{= 4}\)

so \(C = 4\)

which results in satellite's eqn, viz \(s(t)=15t^2-10t+4\)

and you were asked:

"Find the position function, s(t), describing the motion of the object. "

Answer 14

ohhhh ok just that i get it thanks so much!!