Question

Trying to find displacement and total distance traveled. a(t) = 13; v-sub(zero) = -1; 0 < t <2

Answer 1

\(a(t)=13\) right?

Answer 2

Yes

Answer 3

So I get that v(t) = 13t + C and C = -1 at t = 0

Answer 4

pull back one level to get \(v(t)=13t+C\) and since \(v(0)=-1\) it must be \(v(t)=13t-1\)

Answer 5

Then I should go one step further to get s(t)? Then I get 13t^2 - t + C

Answer 6

it is not "C = -1 at t = 0 " but rather \(C=-1\) period

Answer 7

Ok

Answer 8

now like before you have to integrate right?

Answer 9

Yes

Answer 10

That's when I get \[13t ^{2} - t + C\]

Answer 11

not "take the anti derivative" although you will need that

Answer 12

but this time it is a definite integral, not an indefinite one with that \(+C\) out at the end

Answer 13

Why?

Answer 14

good question

Answer 15

you want an actual number out of this right? not just some function whose derivative is \(13x-1\)

Answer 16

Yes

Answer 17

in other words, you want
\[\int_0^213t-1dt\]

Answer 18

the way you do this is by finding an anti derivative of \(13x-1\) namely \(\frac{13}{2}x^2-x\)
evaluate at the upper limit, the lower limit, and subtract
or you could just use plain geometry because all you are looking at is a line

Answer 19

I failed to see the difference between indefinite and definite integration I think.

Answer 20

but if you have a second, we can go back to the original question, why not the \(+C\) and see if i can explain clearly

Answer 21

we could do it in a general case, or work with a specific example

Answer 22

Yes please

Answer 23

ok lets do it with a specific example, lets let \(f(x)=2x+1\)

Answer 24

notice i wrote \(f(x)\) because i am thinking that \(f\) is a function of \(x\)

Answer 25

there are an infinite number of "anti derivatives" for \(f\), anything that looks like
\[F(x)=x^2+x+C\] but they all have to look like that, because if two functions have the same derivative, they can only differ by a constant i.e. it has to be \(x^2+x+number\)

Answer 26

Gotcha

Answer 27

now lets consider this specific function of \(x\)
\[A(x)=\int_2^x(2t+1)dt\] notice that it is a function of \(x\) and not of \(t\)

Answer 28

i.e.
\[A(1)=\int_2^1(2t+1)dt\] and
\[A(3)=\int_2^3(2t+1)dt\] etc

Answer 29

we know from the fundamental theorem of calculus that
\[A'(x)=f(x)=2x+1\]

Answer 30

which means what do we know about \(A\)? it must be \[A(x)=\int_2^x(2t+1)=x^2+x+C\]

but \(A\) is not some general function, it is a specific function, so \(C\) is not some general constant, it is a fixed number we know

Answer 31

Ok

Answer 32

and it is easy to find, because we know one point on the graph of \(A\) namely we know that
\[A(2)=\int_2^2(2t+1)dt=0=2^2+2+C\] telling us that \(C=-(2^2+2)\)

Answer 33

that is why we can write
\[A(x)=\int _2^x(2t+1)dt=x^2+x-6\] a very specific function,
and also why we can compute
\[A(5)=\int_2^5(2t+1)dt\] by taking an anti derivative of \(2x+1\) which is \(F(x)=x^2+x\)
evaluate at the upper limit, evaluate at the lower limit, and subtract
when you evaluate \(F(x)=x^2+x\) at \(2\) you get the \(-6\) that you need to subtract off, i.e. you get \(C=-6\)

Answer 34

well that was rather long winded, but i hope it made some sense

Answer 35

It helped make sense of what was explained in class. It's good to see things from other people

Answer 36

ok good
now we still didn't solve your problem, the displacement and the total distance, but in any case you should be able to do it now
btw you got the anti derivative wrong, it should be \(\frac{13}{2}x-x\)

Answer 37

Yeah. I realized that. I do that a lot it seems. But I was able to figure out the answers, thanks again!

Answer 38

yw