Question

Calculus question

It's question 6b) .

Answer 1

I keep getting 18.55 m/s but apparently the answer is 55 m/s .

Answer 2

How are you computing 18.55?

Answer 3

I evaluate the Integral of the first piece wise function from 0 to 10 and then add 10.

Answer 4

Form 0 to 1 sorry.

Answer 5

From*

Answer 6

couldnt you just integrate the acceleration function to find velocity and then plug in t=1?

Answer 7

That's not correct apparantly.

Answer 8

because velocity is the result from integrating acceleration...

Answer 9

@Jhannybean You'd need an initial condition to do that

Answer 10

Oh i see.

Answer 11

Well We are given at t=0 , v is 10 .

Answer 12

there you go.

Onceyou integrate acceleteration you get v(t).

v(1)=10

Answer 13

acceleration*

Answer 14

Doesn't matter. I get the wrong answer anyways.

Answer 15

>_< LAME.

Answer 16

One sec

Answer 17

Interestingly, If you Integrate from 0 to 10 and THEN add 10 you get 55 but that's the wrong method.

Answer 18

\(\int a(t)\;ds = 9t - 0.045t^{2} + C\) where C is the initial velocity.

Answer 19

Yep...

Answer 20

...and that gives v(t), the velocity function. Do the integration again to produce x(t), the displacement function. Notice how I have entirely avoided any substitution of values. I have a strong desire to get all the functions in front of me.

Answer 21

Any Particular reason for displacement when it specifically asks for Velocity?

Answer 22

Following what should be your standard operating procedure on all such problems, I read he entire problem statement to see what I would need. Once everything is there in front of me, I can answer any number of questions, including those specifically asked.

You may need to consider the sign of your initial velocity. Is it +10 m/s or -10 m/s?

Answer 23

I am going to ASSUME -10 m/s since the Raindrop is falling downwards.

Answer 24

You would think. :-) Just he same, after you have v(t) and you evaluation v(1) to answer part (b), the velocity should be greater than where it started (with the same sign).

If we start with v(0) = -10 m/s
And we get v(1) = -10.2 m/s, then there was positive acceleration and we probably did the right thing.
If we get v(1) = -9.95 m/s, then there was negative acceleration and that is not right for this problem.

I'm just suggesting reasonableness checks along the way. I just made up the numbers.

Answer 25

See, Even if I take the sign into account I am simply not getting 55 m/s .

Answer 26

I conclude the textbook made an error.

Answer 27

Why? \(v(t) = 9t - (9/20)t^{2} + 10\)

\(v(0) = 10\) as desired
What is v(10)?

Answer 28

v of 1 not 10.

Answer 29

v(1) is 18.55 m/s .

Answer 30

Look really hard at a(t). The sign on the acceleration is positive!! Make the initial velocity positive, too.

Answer 31

I have done so. I get 18.55 m/s however.

Answer 32

You didn't answer my question. What is v(10)? I don't much care about the problem statement at this point. I'm trying to make sense of the problem structure.

Answer 33

95.5 m/s .

Answer 34

Wait....

Answer 35

55

Answer 36

55 m/s .

Answer 37

Excellent. We have a 55. If this managed to sneak into v(1), then it's just wrong.

Either the question should say v(10) and 55 is the right answer or the correct answer for v(1) is 18.55.

Answer 38

Ahh it's an error XD .

Answer 39

Thanks!