Question

A particle starts at x(0) = 2. If its velocity is given by v(t) = ln(1+t), find its position at t = 5

Answer 1

Take the integral of velocity from 0 to 5 and then add 2.


\[\int\limits_{0}^{5} \ln(1+t) dt\]

Set u = 1+t, du = dt

So you have

\[\int\limits_{0}^{5}\ln(u)du\]

Which becomes

[uln(u) - u] from 0 to 5

Plug in your values to get

5(ln(5) - 1)

Then add 2, so the position at t =5 is

2 + 5(ln(5) -1)

Answer 2

is there a way to find it without u substitution?

Answer 3

Nope. It's the only way to do that sort of integral unfortunately.

Answer 4

really? oh I thought you can use it with FTC

Answer 5

The FTC states that if you have a function F(x) which is the integral of another function G(x) then the derivative of F(x) is simply G(x). Additionally, the integral of G(x) is just F(b) - F(a).

Answer 6

oh kinda like if it's F'(x) = f(x)?

Answer 7

Pretty much.

Answer 8

ok thanks :)

Answer 9

but wait, isn't u supposed to be ln (u) and du is 1+t? because when i plug in your answer, there is no answer choices for it :/

Answer 10

from your answer it is 5.047 correct?

Answer 11

Ugh, sorry, it's been a long day.

(ulnu - u) from 0 to 5 should be

(1 + t)(ln(1+t) - 1)

so it's actually 6(ln6 -1).

Answer 12

Because u was substituted for 1+t earlier, so we need to exchange it back out.

Answer 13

i keep getting 9.65 but it is not an answer choice? i think there's a step missing or a miscalculation :(

the answer choices are : a) 5.751 b) 7.751 c) 1.792 d) 3.792 e) 3.751

Answer 14

Geezus, I just reread my work and I made another mistake. Sorry D:

So when you put 1+t back in for u

you get

(1+t)(ln(1+t) -1) from 0 to 5. So when you put these values in, you get:

6(ln6 -1) - 1(ln1 -1)

Which gives you the distance traveled. BUT you still have to add 2 to the answer, which will give you the position. So the answer should be B.

Answer 15

i'm confused on the -1ln (1-1) part. i'm sorry lol

Answer 16

@CanadianAsian can you please dissect the derivative section? sorry for asking lol