Question

The acceleration function for a particle moving along a line is a(t)=2t+1 . The initial velocity is v(0)=−12. Then:

The velocity at time t, v(t)= .

The distance traveled during the time interval [0,5] is equal to =

Answer 1

Integrate twice.

Answer 2

which I did and I'm not getting the right answer

Answer 3

After the first integration, what did you get for v?

Answer 4

t^2+t-12

Answer 5

good. Now integrate that with integration limits as provided. What was your answer?

Answer 6

(1/3)t^3+(1/2)t^2-12t

Answer 7

Good. I got 35/6

Answer 8

Sorry, \[\frac{ -35 }{ 6 }\]

Answer 9

says it isn't right because that is what I had gotten first off

Answer 10

Is your answer from a book or an instructor?

Answer 11

its online homework

Answer 12

\[v(t) = \int\limits 2t+1 dt = t^2+t+C \]
Note the initial conditions are v(0)=-12 \[\implies v(t)=t^2+t-12\] that will be your velocity, now you need to find distance, again integrating through \[x(t) = \frac{ t^3 }{ 3 }+\frac{ t^2 }{ 2 }-12t \] we have to note the change of direction when v = 0, that's the trick here.

Answer 13

Astro is right. The particle changes direction when v=0. Negative areas count in the distance moved. I may be getting rusty.

Answer 14

so then what do you do for that?

Answer 15

First, solve for v(t) = 0

Answer 16

The wording is a bit odd, they should write displacement, so you can find all the individual displacements and add them up

Answer 17

Got to go to dinner - will check back, but it looks like Astro has the key.

Answer 18

Astrophysics I need some help

Answer 19

First, when does the particle change direction?

Answer 20

at 0

Answer 21

Right v(t) = 0 so what do you get

Answer 22

-12?

Answer 23

I meant \[v(t) = 0 \implies 0 = t^2+t-12\]

Answer 24

What do you get

Answer 25

t=3 and t=-4

Answer 26

So notice -4 is not in our time interval, so now we'll have two integral from 0 to 3 and 3 to 5

Answer 27

Or you can find individual displacements and add them up

Answer 28

So find x(3)-x(1), x(5)-x(3), and x(1) and add them up

Answer 29

That will give you your total distance

Answer 30

do you put those numbers in the first or second anti-derivative?

Answer 31

You want the distance, so distance

Answer 32

i'm still getting the wrong answer. Sorry I'm not trying to be complicated, I just don't fully understand

Answer 33

Can you show me your work

Answer 34

Well I plugged the (1/3)t^3+(1/2)t^2-12t into the calculator to get those points you said.
5=-5.8333 3=-22.5 1=-11.17, subtracted the 2 intervals you said and then added together what I got from that

Answer 35

Ok one second, so we have v(t) = (t-3)(t+4) correct

Answer 36

Lets do \[0<t<1, ~~~~ 1<t<3,~~~~ 3<t<5\] we can find the individual displacements here

Answer 37

\[x(1)-x(0) = -67/6 \]
\[x(3)-x(1) = -45/2-(-67/6) = -34/3\]
\[x(5)-x(3) = 16/3\] check the math

Answer 38

Your total distance is the absolute value of all that

Answer 39

Hey can you check your initial condition

Answer 40

condition? [0,5]

Answer 41

v(0)

Answer 42

How many attempts do you have btw?

Answer 43

its unlimited attempts

Answer 44

Ok can you try 235/6

Answer 45

that was right

Answer 46

Ok perfect, lets just stick to my first though then, we'll integrate through and get \[- \int\limits_{0}^{3} t^2+t-12 dt - \int\limits_{3}^{5} t^2+t-12dt \]

Answer 47

Because the other way might be very confusing especially since i'm not doing it on paper

Answer 48

\[- \int\limits\limits_{0}^{3} t^2+t-12 dt \color{red}{+} \int\limits\limits_{3}^{5} t^2+t-12dt \]

Answer 49

That's your distance

Answer 50

how did you get those intervals?

Answer 51

Good question, alright so remember when I asked you, to find where the particle changes direction, we need to set v(t) = 0 that gave us 0=(t-3)(t+4) which gives 3, -4 we can neglect the -4 since it is not in our interval [0,5], but the 3 here tells us when the particle changes direction you can even try plugging in the values in v(t) to check your self, you will see 0<t<3 we get v<0, then 3<t<5 we get v>0 and we have that negative in front of the first integral because other wise we will get a negative value, but here we want the distance so we are taking the positive values or essentially the absolute value, you can note \[\int\limits_{a}^{b} |v(t)|dt = distance\]and \[\int\limits_{a}^{b} v(t) dt = displacement\]

Answer 52

ok thanks. I understand now

Answer 53

Yeah haha sorry about all the mess earlier, should've just went with my first though

Answer 54

thought*