Question

The acceleration of an object is given by the equation a(t)=6t^3-3t^2+2 m/s^2
Determine its velocity and position after 1 seconds, assuming s(0)=4m and v(0)= 0 m/s

After 1 seconds, the object's velocity is______m/s
and its position is___________m

Answer 1

find the anti derivative of \[6t^3-3t^2+2\]

Answer 2

Velocity is the anti-derivative of acceleration.
Position is the anti-derivative of velocity.

Answer 3

your answer will have a C in it, which you can solve for by setting \(v(0)=0\) and seeing that C must be zero as well

Answer 4

3x^4/2-x^3+2x+c

Answer 5

i can walk you through the steps if you like, it is all using
\[\int t^n=\frac{t^{n+1}}{n+1}\]

Answer 6

yes

Answer 7

3t^4/2-t^3+2t+c

Answer 8

\[\frac{3}{2}t^4-t^3+2t+
c\]

Answer 9

and since \(v(0)=0=c\) you also have \(c=0\) making it
\[v(t)=\frac{3}{2}t^4-t^3+2t\]

Answer 10

okay..

Answer 11

then the first question asks for \(v(1)\) which is now real easy, replace \(t\) by \(1\)

Answer 12

then find the anti derivative a second time to get \(s(t)\)

Answer 13

so 5/2 will be v(1)

Answer 14

and then 101/20

Answer 15

i am not sure what you mean by "then 101/20"

Answer 16

thanks!

Answer 17

can you help me with another problem?

Answer 18

\(v(1)=\frac{5}{2}\) looks good to me

Answer 19

but i did not find \(s(t)\) did you?

Answer 20

sure

Answer 21

i got that for the second part
i found the second dreviitve and i s(1) and c=4, it came up as the right answer

Answer 22

although i am not sure you finished with this one
did you find \(s(t)=\frac{3}{10}t^5-\frac{t^4}{4}+t^2+4\)?

Answer 23

The marginal revenue for x items in dollars is given by r'(x)=-2.5x+7

determine and the revenue function and demand function

Answer 24

oh i see you are right

Answer 25

can you help me with the problem i posted please

Answer 26

i am not sure about this one
a quick google search tells me marginal revenue is \[R(x)=P(x)x\] does that look familiar?

Answer 27

yes but im not sure if it has to do with this

Answer 28

we have
\[R'(x)=-2.5x+7\] so
\[R(x)=-1.25x^2+7x\]

Answer 29

if we write
\[-1.25x^2+7=(-1.25x+7)x\] it looks like \[P(x)x\]

Answer 30

i am not exactly sure either, sorry
it looks like you want two things, revenue function and demand function
maybe i can find it in a book

Answer 31

thanks though
im going to repost it to see if anyone can help

Answer 32

ok

Answer 33

it looks like "revenue function" is \(R\) so in this case it is
\[R(x)=-1.25x^2+7x\] and demand function is \(p(x)=-1.25x+7\) since
\[R(x)=xP(x)\]