Question

The velocity function, in feet per second, is given for a particle moving along a straight line.
v(t) = 5t − 4, 0 ≤ t ≤ 3
(a) Find the displacement.
(b) Find the total distance that the particle travels over the given interval.

Answer 1

Hint: \[v(t) = \frac{ dx(t) }{ dt } \]
You'll have to integrate this over the given interval

Answer 2

I got all the way through it but not the right answer. I got -3/4 for my displacement

Answer 3

I did integrate it. I'm just not coming up with the correct answer for some reason

Answer 4

You may have made an arithmetic mistake, because -3/4 doesn't sound right

Answer 5

where did you get the 9t?
I got

Answer 6

When you integrate a constant, you get at+C so \[\int\limits 9 dt = 9t+C\]

Answer 7

you don't need a constant it's on a definite integral

Answer 8

I know, I'm just explaining the 9t

Answer 9

\[\int\limits_{0}^{3} 9 dt \implies 9t|_0^3\]

Answer 10

yes! okay awesome I fixed part a with 21/2

Answer 11

some handy rules, just in case/ review

Answer 12

AWESOME that's so helpful thank you

Answer 13

Yw, so what did you get for your displacement now?

Answer 14

21/2

Answer 15

hmmm it told me it was correct

Answer 16

but I can't get part b

Answer 17

Oh sorry I kept using 9 when there is a 4 haha :)

Answer 18

hahaha okay :) now I got 87/10 for part b & they counted it wrong

Answer 19

Let me just fix that to show it again \[\int\limits\limits_{0}^{3} (5t-4) dt \implies (\frac{ 5t^2 }{ 2 }-4t)|_0^3\] ah there we go

Answer 20

The distance is a bit tricky, because we have to find the total interval so we have \[5t-4 = 0 \implies t = \frac{ 4 }{ 5 }\] so our integral should be \[\int\limits_{0}^{4/5} (5t-4)dt + \int\limits_{4/5}^{3} (5t-4)dt\] and note that distance is a scalar quantity so you will have to use absolute values for your integrals

Answer 21

At t =4/5 is really just when it changes the direction

Answer 22

OH oh my gosh that makes so much more sense

Answer 23

i got 73/10 this time.. still incorrect any suggestions?

Answer 24

\[|\int\limits_{0}^{4/5} (5t-4)dt| \implies |(\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} |\] the integral is the same your interval is just changed so here is the first one

Answer 25

The |...| represent absolute values

Answer 26

Yes! I did that

Answer 27

Ok what did you get for this integral?

Answer 28

\[(4t-\frac{ 5t ^{2} }{ 2 }) Integral (0, 4/5) + (\frac{ 5t ^{2} }{ 2 } - 4t) \int\limits (3, 4/5)\]

Answer 29

Why did you write \[4t-5t^2/2\]

Answer 30

because that's on the integral (0, 4/5)? maybe not?

Answer 31

Noo, why are you changing the integrand? It's the same we are just seeing where the velocity is negative and positive

Answer 32

okay so do I set the integral negative? hahaha I'm so sorry

Answer 33

No, we are finding the distance, which is the magnitude it cannot be negative

Answer 34

so we're just finding the velocity which means just plugging in the integral? & keeping the integral the same?

Answer 35

Nooo, we're finding the distance, and the reason we have split it into two integrals is because the integral 0 to 3 does not account for the negative and positive path of the particle, that is just the displacement, but for part b we want distance, that's total area travelled, so we have to find the change in velocity which is at \[t=4/5\] so we go from 0 to 4/5 and we add the positive area 4/5 to 3 for the total distance...I hope that makes sense now

Answer 36

v(t) = 5t-4 is a straight line, the line does not change, but the path does

Answer 37

If you meant put the first integral as \[- \int\limits_{0}^{4/5} (5t-4)dt\] that would work

Answer 38

OH okay let me try!

Answer 39

would my second integral be \[\int\limits_{0}^{3} (5t-4)\]

Answer 40

Noooo haha, I have already set up the integrals for you, hmm ok, I think you're not understanding the concept

Answer 41

I feel like I'm losing it more and more hahaha okay let me look this over again

Answer 42

\[\left| \int\limits_{0}^{4/5}(5t-4)dt \right|+\left| \int\limits_{0}^{4/5} (5t-4)dt\right|\] that's all there is!

Answer 43

HAHAHAHAHA OH MY GOSH

Answer 44

any ideas why it's not taking my answer of 3.2? sometimes it's picky

Answer 45

That's not right

Answer 46

Lets do this one at a time \[\int\limits_{0}^{4/5} (5t-4)dt\] what does this integral =? Remember to include absolute value

Answer 47

I got -1.6 but of course the absolute value is 1.6

Answer 48

the integral is \[\frac{ 5t ^{2} }{ 2 }-4t \int\limits (0,4/5)\]

Answer 49

Why is your notation like that, \[\int\limits\limits_{0}^{4/5} (5t-4)dt = (\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} = \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right) ^2}{ 2 }-4(\frac{ 4 }{ 5 }) \right)-0 = |\frac{ 5 \times 16}{ 2 \times 25}-\frac{ 16 }{ 5 }| = \frac{ 8 }{ 5 }\] so that's good, that is 1.6

Answer 50

Now lets do the other integral \[\int\limits_{4/5}^{3} (5t-4)dt = \left( \frac{ 5t^2 }{ 2 }-4t \right)|_{4/5}^{3}\]

Answer 51

it's normally not like that I just struggle with plugging it into a computer, my apologies

Answer 52

\[\frac{ 121 }{ 10 }\] this integral gives the above, you can check

Answer 53

Now we add the two integrals \[\frac{ 8 }{ 5 } + \frac{ 121 }{ 10 } = \frac{ 137 }{ 10 }\]

Answer 54

when I plug 3 in, i get \[\frac{ 45 }{ 2 } - 12=21/2\] ...what is going on

Answer 55

\[\left( \frac{ 5(3)^2 }{ 2 } -4(3)\right) - \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right)^2 }{ 2 } -4(\frac{ 4 }{ 5 })\right)\] this is the result of the second integral, you're forgetting about \[\frac{ 4 }{ 5 }\]

Answer 56

yes so this would be 21/2 - 8/5 = 89/10

Answer 57

No, it should be +8/5, the second term gives -8/5 remember? And you're already subtracting using FTC so a (-)(-) = +

Answer 58

We figured this out with the first integral

Answer 59

ohhhhhhhhhhhhhh