Finding the total distance traveled by the body at interval

[GustX]

Homework Statement

At time t, the position of a body moving along the s-axis is
$s= t^3 -12t^2 + 36t m(meters)$
Find the total distance traveled by the body from t = 0 to t = 3.

Homework Equations

Derivatives

The Attempt at a Solution

I got the derivative which is
3t^2 - 24t + 36(meters)
I set it equal to zero and got

t = sqrt[-4]

which wouldn't work out

I've also tried inputting the derivative into the quadratic formula and got
$t = 4$
$t = 2$
idk if that was neccesaryi tried inputting all the t values from interval 0-3 and got the following results
$0 -> 0$
$1 -> 25$
$2 -> 32$
$3 -> 27$

Trying to mimic the book I took the differences between each interval result
$(0 and 25) -> 25$
$(25 and 32) -> 7$
$(32 and 27) -> 5$
$25 + 7 + 5 = 37<br />$
but that isn't the answer on the sheet either which is 59m

 

[LCKurtz]

Homework Statement

At time t, the position of a body moving along the s-axis is
$s= t^3 -12t^2 + 36t m(meters)$
Find the total distance traveled by the body from t = 0 to t = 3.

Homework Equations

Derivatives

The Attempt at a Solution

I got the derivative which is
3t^2 - 24t + 36(meters)
I set it equal to zero and got

t = sqrt[-4]

which wouldn't work out

I've also tried inputting the derivative into the quadratic formula and got
$t = 4$
$t = 2$
idk if that was neccesaryi tried inputting all the t values from interval 0-3 and got the following results
$0 -> 0$
$1 -> 25$
$2 -> 32$
$3 -> 27$

Trying to mimic the book I took the differences between each interval result
$(0 and 25) -> 25$
$(25 and 32) -> 7$
$(32 and 27) -> 5$
$25 + 7 + 5 = 37<br />$
but that isn't the answer on the sheet either which is 59m

You need to show more work. It matters where the derivative (velocity) is positive or negative. If the object switches direction and doubles back on itself, the difference in position is not the total distance travelled. Start by showing us where on the interval the derivative is positive or negative. Look at the signs of the factors of the derivative to figure that out.

[Edit, added] Looking again, I think your text is incorrect and your answer of 37 is correct. But your method isn't correct because there is no reason to just restrict yourself to integers when evaluating the position.

 

[GustX]

You need to show more work. It matters where the derivative (velocity) is positive or negative. If the object switches direction and doubles back on itself, the difference in position is not the total distance travelled. Start by showing us where on the interval the derivative is positive or negative. Look at the signs of the factors of the derivative to figure that out.

[Edit, added] Looking again, I think your text is incorrect and your answer of 37 is correct. But your method isn't correct because there is no reason to just restrict yourself to integers when evaluating the position.

when factored its

3(t-2)(t-4)

you mean the signs atm or after i find the zeros?

Since the quadratic gave me
t =2
t = 4
it should change in these two points but only 1 (t =2) is in the interval so only 1 change matters
but when i input t values t >= 2 the sign doesn't change from the previous s(t) values which were t < 2
I don't think my answer is correct since it isn't the same as the sheet which says 59m,I put in non integers now

0.5 -> 15.125
1.5 -> 30.375
2.5 -> 30.625

 

[LCKurtz]

when factored its

3(t-2)(t-4)

That is why you need to show your work. When what is factored? I don't get those factors anywhere.

you mean the signs atm or after i find the zeros?

What does atm mean?

Since the quadratic gave me
t =2
t = 4
it should change in these two points but only 1 (t =2) is in the interval so only 1 change matters

What should change?

but when i input t values t >= 2 the sign doesn't change from the previous s(t) values which were t < 2
I don't think my answer is correct since it isn't the same as the sheet which says 59m,I put in non integers now

s(t) is positive on the whole interval [0,3]. Why do you expect its sign to change? You seem very confused about this.

0.5 -> 15.125
1.5 -> 30.375
2.5 -> 30.625

It does no good to type in random numbers. You need to think about the velocity and when it is positive and negative. That will give you the important values of $t$ where the object changes direction. Look again at my suggestion in post #2 in the paragraph before the edit. Also note my edit in post #2 where I indicate the answer of 59 that is given is incorrect.

 

[GustX]

That is why you need to show your work. When what is factored? I don't get those factors anywhere.

Ok, I wrote a number wrong when i factored just then but original problem so it is
$3(t-2)(t-6)$

What does atm mean?

if whether I should look at the signs of the factors of the derivative to find out where on the interval its positive or negative

What should change?

positive and negative?

s(t) is positive on the whole interval [0,3]. Why do you expect its sign to change? You seem very confused about this.

It does no good to type in random numbers. You need to think about the velocity and when it is positive and negative. That will give you the important values of $t$ where the object changes direction. Look again at my suggestion in post #2 in the paragraph before the edit. Also note my edit in post #2 where I indicate the answer of 59 that is given is incorrect.

Weird that it would be incorrect but maybe the prof made a mistake.

 

[LCKurtz]

Ok, I wrote a number wrong when i factored just then but original problem so it is
$3(t-2)(t-6)$

You keep saying "it is" but your expression needs an equal sign and something on the left side to tell what "it" is.

Weird that it would be incorrect but maybe the prof made a mistake.

Also weird that you don't seem to be interested in actually analyzing the problem. If "it" is the derivative of $s(t)$ and you figure out where it is positive or negative, what are you going to do with that knowledge to actually solve the problem correctly?