Question

Calculus Help
Walk through.

Answer 1

The graph shows the acceleration function
Find the function represented for that graph
Integrate to find the velocity function and use the condition to find what the constant of integration should be
then integrate the velocity function to find the position function and use the condition to find this constant of integration

Answer 2

Freckles has made some good suggestions here.

First of all, find the slope of the line that represents the acceleration as a function of time t. I see from the graph that the "rise" is actually negative, and is -10 ft/(sec^2) for a "run" of 10 sec. The vertical intercept is 10 ft/(sec^2). Find the equation of this line.

Answer 3

y=-1x+10. I found th slope to be -1. Then plugged it into the slope of a line equation.

Answer 4

@mathmale

Answer 5

All right. Before we proceed: are you in a calculus course? are you familiar with definite integrals?

Answer 6

Yes.

Answer 7

All right. Good. a(t)=[10-t] ft/(sec^2).
Integrate this to obtain an equation for velocity.

Answer 8

Wait what

Answer 9

acceleration is the derivative of velocity. velocity is therefore the integral of acceleration.

Answer 10

By starting with the formula for acc'n and integrating, we can obtain a formula for the velocity. By integrating this velocity formula, we can obtain a formula for the displacement or distance traveled.

Answer 11

So the y=-1x+10 is the velocity or acceleration formula?

Answer 12

That is the acceleration right?

Answer 13

What do you think? What does the graph represent?

Answer 14

Acceleration.

Answer 15

Right, and we seem to agree that the acc'n, a(t), is -t+10 ft/(sec^2). Right?

Answer 16

I thought it was a(t) = -1x+10 ?

Answer 17

Wouldn't it make sense to use the independent variable t (for time)? At this level of math, you can't mix x and t; that happens in differential equations.

Answer 18

OH, right. Agreed.

Answer 19

\[\int\limits_{0}^{30} -t+10 dt\]

Answer 20

would that be how i integrate it?

Answer 21

What does the question ask for? I ask you this because of your limits, 0 to 30. Please enclose your -t+10 inside parentheses as you write the definite integral.

Answer 22

It wants to know the position of the car at time 20 sec.

Answer 23

Yes. So, there's no reason to use 30 as your upper limit.

Also, integrating a(t) produces a formula for velocity, v(t), not for position, s(t).

Please enclose -t+10 inside parentheses and write the integral that will produce an equation for v(t).

Answer 24

\[v(t) = \int\limits_{0}^{20} (-t+10) dt\]

Answer 25

You're not ready to use limits of integration yet.

We begin with a(t), obtaining an indefinite integral:\[v(t) = \int\limits\limits_{}^{} (-t+10) dt\]

Answer 26

Complete the integration, please. show your result.

Answer 27

\[\frac{ t^2 }{ 2 }+10t+C\]

Answer 28

Right, except that we must keep that - sign which is part of -t+10, and your result should be labeled v(t).\[-\frac{ t^2 }{ 2 }+10t+C=v(t)\]

Answer 29

RIght.

Answer 30

What is the initial velocity / speed? Subst. that for C in your equation, above.

Answer 31

It says 0?

Answer 32

Yes. then, our equation for velocity is \[v(t)=-\frac{ t^2 }{ 2 }+10t\]

Answer 33

Pls integrate this with respect to time, t, including the required constant of integration and the label s(t).

Answer 34

\[s(t) = \frac{ t^2 }{ 6 }+5t^2+C\]

Answer 35

I forgot the negative, My bad. -t^2/6

Answer 36

Why the final power 2 instead of 3?

Answer 37

Crap, I wrote it wrong. it is 3.

Answer 38

\[s(t) = -\frac{ t^3 }{ 6 }+5t^2+C\]

Answer 39

What is the initial position, s(0)?

Answer 40

=1

Answer 41

Wow. I meant it is just C.

Answer 42

Please reread the original question, so that you can answer mine about the initial position.

Answer 43

Oh sorry, it is 10.

Answer 44

at s(0) the position is 10.

Answer 45

So, replace your C with 10 feet.

Answer 46

\[s(t) = -\frac{ t^3 }{ 6 }+5t^2+10 feet\]

Answer 47

Now substitute t=20 sec. s(20)=?

Answer 48

s(20) = 676.66 ?

Answer 49

same result as mine. Did we both take that - sign into account?

Answer 50

yes.

Answer 51

Seems to me that our result is correct, then. We used all of the given info, incl. the initial position and the initial velocity. I'd interpret our result to state that "The net distance traveled in the first 20 sec was 676.7 feet." You OK with this?

Answer 52

How is that answer even possible?

Answer 53

Let me turn that question around and ask you why you feel it's not possible?

Answer 54

Well we did all the steps right, so Im okay with the answer. I dont know why it wouldnt be possible. so. 676.667 ft/sec^2 ?

Answer 55

I'd round that off to 676.7 feet (NOT 676.7 ft/(sec^2). We're measuring displacement, not acceleration.

Answer 56

It says round to 3 decimal places. and okay.