Question

HELP WITH CALC! PROBLEM ATTACHED!

Answer 1

v(t) should be s'(t)

so ds/dt = v(t)


s(t) is the integral of v(t)

remember when you have an indefinite integral, you get + C at the end.

C is found by plugging in s(3) = 0

Answer 2

so find the integral of that equation thats al they're asking for?

Answer 3

yes.

since it'll come in the form s(t) + C, you have to find C and include it in your final answer

Answer 4

so plug in 0 in the original equation or in the integral?

Answer 5

\[\frac{ 5t^3 }{ 3 }-2t^2\] is my integral so is that ds/dt?

Answer 6

\[v(t) = \frac{ ds }{ dt }\]so\[s(t) = \int\limits_{}^{} v(t) dt\]

Answer 7

no work required for part A. you just need to know that the derivative of position is velocity.

ds/dt = 5t^2 - 4t

you forgot the very important constant
\[s(t) = \frac{ 5t^3 }{ 3 } - 2t^2 + C\]

they gave you an initial condition to solve for C. this is when t = 3, s = 0

or s(3) = 0

plugging that in s(t)

\[0 = \frac{ 5(3)^3 }{ 3 } - 2(3)^2 + C\]\[C = 27\]final/real answer:\[s(t) = \frac{ 5t^3 }{ 3 } - 2t^2 + 27\]

Answer 8

oh ok it makes so much more sense when you explain it in English! thank you so much!

Answer 9

hehe :) glad i could help ^_^

Answer 10

wait its telling me that s(t) is wrong..

Answer 11

i disagree. should be right ^_^

Answer 12

please copy/paste what you entered in the box

Answer 13

oh. my mistake. went too fast C = - 27

Answer 14

haha now its right sorry for the confusion

Answer 15

lol its my fault :P