Question

Calc AB Question
The function v(t) = t^2-3t-10 is the velocity in m/sec for a particle moving along the x-axis, where t is measured in seconds (t >/= 0).

a. Determine when the particle is moving to the right, to the left, and stopped on the interval 0
12 years ago

Answer 1

I think for the b. we need to do \[\int\limits_{0}^{7}v(t)\]

@Kainui am i correct ?

Answer 2

Yeah, but remember the dt!

Remember, distance=rate*time so literally an infinitesimal bit of movement will be:

dx=v(t)*dt

add up all the infinite number of infinitesimally small distances and you get:\[distance = \int\limits_{0}^{7}v(t)dt\]

Answer 3

:) I missed the dt.

Answer 4

All good, I just wanted to explain sort of the reasoning behind it to @MathGeekinProgress

Answer 5

For part a) take the derivative of v(t), set it equal to 0 and solve for t to see where the velocity changes sign.

Answer 6

Oh I know about the dt but honestly, where do I start? What about part a?

Answer 7

Thanks @whpalmer4 for part a haha

Answer 8

The derivative ov v(t) is 2t-t

Answer 9

Check your work...

Answer 10

Well what makes v(t) different when it's moving to the left vs the right? How about when it's stopped?

Answer 11

Wouldn't that be the same as a(t)?

Answer 12

Woops. 2t -3*

Answer 13

The derivative of v(t) is a(t), yes.

Answer 14

you know what, it's late here, scratch what I said...

Answer 15

So 2t-3=0 is t=3/2

Answer 16

What???

Answer 17

No warranty of correctness applies :-)

Answer 18

You're in good hands with these guys, I'm going to bed!

Answer 19

How do I know where the sign changes?

Answer 20

-_-

Answer 21

a) v(t) < 0, the particle moves left
v(t) > 0, the particle moves right

b) ∫v(t) dt, from 0 to 7

c) remember, distance isn't the same as displacement.
distance = ∫│v(t) │dt, from 0 to 7

I will let you do theh math

Answer 22

@MathGeekinProgress Yeah sourwing's on the right track.

Answer 23

For part a, if v(t) = 0, then we know the particle is stopped, correct?

Answer 24

correct

Answer 25

So for part a, the particle is moving towards the riight, correct?

Answer 26

v(t) = 0 = t^2-3t-10 = (t - 5) (t + 2),
t = 5 (exclude - 2)

so you have to check the interval (0,5) and (5,7) for the sign of v(t)

Answer 27

I dont get it..

Answer 28

For Part B I got -174/6. For Part C I got 1547/3.

Answer 29

well, at t = 5 is when the particle stopped. Everywhere else is the particle either moving left of right. Like I said before, v(t) > 0 means the particle moves to the right and v(t) < 0 means the particle moves left

Answer 30

B) it's -175/6 (typo?)

C) it's 125/2

Answer 31

Yeah typo

Answer 32

How did you get 125/2?

Answer 33

how did you get 1547/3?

Answer 34

I did the same exact thing for part C as I did for part B. The only difference was that for part B I used t^2-3t-10. For part C I used t^2+3t+10

Answer 35

|t^2 - 3t - 1| isn't the same as t^3 + 3t + 10

Answer 36

t^2 - 3t - 10 if t^2 - 3t - 10 > 0
|t^2 - 3t - 10| =
-(t^2 - 3t - 10) if t^2 - 3t - 10 <= 0

Answer 37

Ohhhhhh. So it's gonna be - t^2 + 3t + 10 because v(t) = 0

Answer 38

in other words:

t^2 - 3t - 10 if 5 < t < 7
|t^2 - 3t - 10| =
-(t^2 - 3t - 10) if 0 < t < 5

Answer 39

\[distance = \int\limits_{0}^{5}-(t^2-3t-10)dt + \int\limits_{5}^{7} (t^2-3t-10)dt\]

Answer 40

Ohhhhhhh

Answer 41

let me solve this

Answer 42

which turns out to be 275/6 + 50/3 = 125/2

Answer 43

Yup I got that!

Answer 44

Now what about part a?

Answer 45

well, where is v(t) < 0? v(t) > 0?