Question

S''(t)=5+4t-2t^2 initial velocity v(0)=3 m/s initial displacement s(0(=10m find position after t seconds

Answer 1

@zepdrix @satellite73

Answer 2

@mathmale

Answer 3

S''(t) is a(t), but they want v(t) and s(t)

Answer 4

Sorry it's A''(t) for beginning

Answer 5

Okay so anti differentiate

Answer 6

Right, it will then equal v(t), since the derivative of v(t) is a(t)

Answer 7

Okayso
V'(t)=5t+2t^2-t^3?

Answer 8

Since S''(t) = a(t) where a is the acceleration you need to integrate to get the velocity function.

Answer 9

I would check it out

Does taking the derivative of your v(t) give you their a(t)?

Answer 10

Yes
\[\frac{ dv(t) }{ dt } = a(t)\]

Answer 11

So the velocity will be
\[v(t) = \int\limits_{}^{} 5+4t -2t ^{2}\]

Answer 12

\[v(t) = 5t-2t^2-t^3\]

So a(t) =

\[5-4t-3t^2\], which is different from theirs

Answer 13

Oh so I was wrong? :O oops

Answer 14

Would it be -2/3t^3 and plus I the middle term?

Answer 15

Right, it would just be 5-4t-2/3t^3

Answer 16

I mean 5t-2t^2-2/3t^3

Answer 17

Hmmmmm it's plus 4t in the middle?

Answer 18

Mb, Im mixing numbers up lol

Answer 19

It's okay :)

Answer 20

Do it again?

Answer 21

Right, let's do it now. But we forgot 1 thing


The +C

Answer 22

Omg I always forget the c!!!!

Answer 23

So it's

\[v(t) = -\frac{ 2 }{ 3 } t^3-2t^2+5t+C\]

They give that v(0) = 3. So let's plug in 0 for t

Answer 24

Okay

Answer 25

\[\frac{ 2 }{ 3 }(0)^3 - 2(0)^+5(0)+C = 3\]

Answer 26

Or something like that lol

Answer 27

So 0+c=3

Answer 28

C=3

Answer 29

Right, so now we have

\[v(t)=-\frac{ 2 }{ 3 }t^3+2t^2+5t+3\]

They want s(t) also, so let's integrate again

Answer 30

I have to go, but do the exact same thing we did for v(t), and that function is your answer

Remember the +C

Answer 31

s(t)=ugh this is getting to hard blah!

Answer 32

Thank you so much! :)

Answer 33

We can integrate this one together if you like

Answer 34

If you have to go it's okay

Answer 35

Nah it's cool. It won't take long.

Since it's adding/subtracting we can separate each part

Whats the integral of -2/3t^3?

2t^2?

5t?

3?

Which one of those can you do right now?

Answer 36

3 I can do right now, 3t

Answer 37

Alright, that one is good

For the rest, it's basically the power rule backwards

So for 5t... thats also \[\frac{ 5t^1 }{ 1 } \]

\[\frac{ 5t ^{1+1} }{1+1 }\]

Answer 38

1 + 1 is the exponent on the t +1