An object at a position s(0) = 4 begins to move at a velocity v(t) = e^(t/5). Find its position at t = 10.

Question

anonymous · Answer

okay list all important information

anonymous · Answer

s(0) = 4 what?

anonymous · Answer

that's a strange velocity laughing out loud

anonymous · Answer

$$s(t) = \int\limits v(t)dt$$

anonymous · Answer

$$= 5e^{\frac{ t }{ 5 }} + c$$

anonymous · Answer

use that 4 for your c

anonymous · Answer

then solve for 
s(10)

anonymous · Answer

I hope @sourwing approves of my solution

anonymous · Answer

it's a strange unitless physical application

anonymous · Answer

these are the answer choices: 
        e2 - 1
	e2 - 5
	5e2 - 1
	5e2 - 5

abb0t · Answer

0<t<10

abb0t · Answer

is what you should be integrating.

anonymous · Answer

what about c, then?

abb0t · Answer

There is no constant, you're integrating from $\sf \color{red}{ 0<t<10}$

anonymous · Answer

so the s(0)=4 was unnecessary info?

abb0t · Answer

i didn't even see that. Uh, so you have $\sf \color{red}{4=5+C}$, solve for $\sf \color{blue}{C}$

abb0t · Answer

But your answer choices are wrong, if that is the case.

abb0t · Answer

It would not be 5e$\sf ^2-1$

abb0t · Answer

Where is the e$^2$ coming from?!?!?!

anonymous · Answer

so the answer is 5e^2−1. because you solve the integral, then find c which is -1.

anonymous · Answer

where is 5 coming from when solving C?

anonymous · Answer

that was incorrect. i was solving the definite integral someone gave and plugged that it. ignore that whole 5e^2−5  thing whoops.

anonymous · Answer

c=4 is not an unnecessary information
s(0) means that is your initial position, that is a t=0 you are at 4

anonymous · Answer

the derivative of position function is velocity and the anti-derivative (integral) of velocity is position function 
we took the integral of velocity and in place of C we put the initial position

anonymous · Answer

wait how is c=4? 5e^t/5 is the anti derivative of the given equation, right? if you plug t=0 into that anti derivative, wouldn't you get 5? e^0=1. therefore c=-1?

anonymous · Answer

$$s(t)=\int\limits v(t)dt= \int\limits e^{\frac{ t }{ 5 }}dt=5e^{\frac{ t }{ 5 }}+C$$
to solve for C
$$s(0)=5e^{\frac{ t }{ 5 }}+C=4$$
looking at the exponent $$\frac{ 0 }{ 5} = 0  \rightarrow 5e^0 = 5*1=5$$
5+C = 4
C = -1

evaluate at s(10)
meaning plug 10 in everywhere you see a t

anonymous · Answer

so the answer is 5e^2−1

anonymous · Answer

$$s(t)=5e^{\frac{ t }{ 5 }}-1$$

anonymous · Answer

okay thank you!

anonymous · Answer

the answer has been given 
you just need to apply yourself a little bit