Question

Calculus help, picture below, just part C and D

Answer 1

Which part are you specifically stuck on?

Answer 2

i was only able to do part A

Answer 3

What is preventing you from answering part B?

Answer 4

I don't understand how to do it

Answer 5

Define speed for me.

Answer 6

but I know the answer is -2

Answer 7

Define speed for me.

Answer 8

distance over time

Answer 9

ohh nvm i got part B now

Answer 10

What's catching you up on part C?

Answer 11

idk how to find t? and what to do with (0,4)

Answer 12

Think about what "Change of Direction" means..what are they trying to tell you about the particle at the instant it changes direction? What are some properties a particle has when it's going one way versus the opposite direction?

Answer 13

well it can change to negative

Answer 14

So what does that mean when you have a positive changing to negative or a negative changing to positive?

Answer 15

derivative?

Answer 16

to answer these questions, you need to know that if you have a function that gives position as a function of time
x(t)
then velocity is \( v= \frac{dx}{dt} \)
and acceleration is \( \frac{dv}{dt}= \frac{d^2x}{dt^2} \)

Answer 17

the question states
\[ x(t) = 7t-4t^2+\int_0^t s^2 ds \]
I would simplify this by evaluating the integral.
can you do that ?

Answer 18

would i take out the t and have t(7+4t)?

Answer 19

no, not the first two terms. can you simplify
\[ \int_0^t s^2\ ds \]

Answer 20

how?

Answer 21

do you know how to integrate \( \int x^2 dx \) ?

Answer 22

have you studied integration ?

Answer 23

yea but i hardly remember

Answer 24

see http://www.khanacademy.org/math/calculus/integral-calculus/indefinite_integrals/v/indefinite-integrals-of-x-raised-to-a-power

for the details.

you should get
\[ \int_0^t s^2 \ ds = \frac{1}{3}s^3 |_0^t=\frac{t^3}{3} \]

Answer 25

your equation is now
\[ x(t)= 7t-4t^2+ \frac{1}{3}t^3 \]

Answer 26

oh yea i did something like that for part b

Answer 27

what do you get for part (a)? position and velocity at t= 3 ?

Answer 28

-8

Answer 29

position = ? velocity = ?

Answer 30

velocity=-8 and position=-6

Answer 31

ok, and to find the velocity, I assume you found dx/dt ?

Answer 32

yes

Answer 33

for (b), I don't know if they are asking for average speed or instantaneous speed.
I would *guess* they want instantaneous speed = | instantaneous velocity|

Answer 34

yea i did part be i got -2

Answer 35

How did you get -2?

Answer 36

v=distance/time so i used -6 from part A and divided it by 3 since t=3 and i got -2

Answer 37

that is average speed from t=0 to t=3. Are you sure that is what they want ?

Answer 38

the other possible answer is speed=8 (this is instantaneous speed)

also, they ask, is the particle speeding up or slowing down at t=3

Answer 39

i put decreasing since its negative

Answer 40

I would look at the second derivative. Did you find it, and evaluate it at t=3 ?

Answer 41

no, i just know that its -2 cause i have the answer sheet

Answer 42

What did you get for dx/dt ?

Answer 43

Speed cannot be negative.

Answer 44

idk thats what it says

Answer 45

and for part c it says t=1, but idk how to get that

Answer 46

Future makes a good point.... speed does not have a direction (i.e. it is a positive number).
If you found the 2nd derivative, at t=3 the acceleration (change in velocity) is -2. The velocity is -8. So you are heading to the left (more negative) and the acceleration is in the same direction... in other words the velocity is becoming more negative (going faster toward the left). I would say the speed is increasing.

Answer 47

for part (3), imagine the particle is moving to the left at some speed. If it turns around and goes in the other direction, it must slow down... all the way to zero, then speed up in the opposite direction. The idea is *the spot where the velocity is zero is where the particle changes direction*

Answer 48

What did you get for dx/dt ?

Answer 49

7-8t+t^2

Answer 50

now use the idea that when the velocity is zero, that is where the particle changes direction. In other words, solve for t:
7-8t+t^2 = 0

Answer 51

to solve
t^2 -8t + 7= 0
factor it.

Answer 52

(t-1)(t-7)=0 t=1, t=7

Answer 53

but t=7 is not in interval

Answer 54

that means it changes direction at time t=1 and at time t=7
but the problem restricts the time to be 0≤ t ≤ 4
so t=1 is the answer.

Answer 55

ok with part c ?

Answer 56

yes

Answer 57

for part d. we know a few things.
at time t=0 , the position is 0
it moves in the same direction up to time 1.
It turns around, and moves in the opposite direction up to (and beyond) time 4

can you find the furthest to the right and the furthest to the left the particle gets in the (0,4) time interval ?

Answer 58

farthest right at t=1 and farthest left at t=4?

Answer 59

actually, they only want the times... which we know

Answer 60

How do we know the furthest right is at t=1 and not at t=4 ?

Answer 61

Idk my answer booklet says thats

Answer 62

can you think of how to show t=1 the particle is at the right ?
one way is find x(1) and x(4)
if x(1) is to the right of x(4), then t=1 is the right most position

or you could think like this:
we know from part (a) that at t=3 the particle is moving -8 (that is to the left)
so at t=4 it must be as far left as it is going to get (in the interval 0 to 4)

also, at t=1, it change direction, so it must having been moving to the right from t=0 up to t=1. at t=1 it is as far right as it gets, before turning around and heading left.

Answer 63

ooh okayy