The velocity problem...

L

Lizzie

Full Member
Joined
Sep 8, 2005
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317
OK, I had a similar question before.

The Problem:
The velocity of a particle moving along a line is \(\displaystyle v(t)=t^3-t\) meters per second. Find the distance traveled in meters during the time interval \(\displaystyle {0}\le{t}\le{2}\). (It helps to sketch a graph of the function before answering the question.)

What I've done:
\(\displaystyle \L\int^2_0{}(t^3-t)dt\)
. . .
\(\displaystyle \L\int^1_0{}(\frac{t^4}{4}-\frac{t^2}{2}) + \int^2_1{}(\frac{t^4}{4}-\frac{t^2}{2})\)
. . .
\(\displaystyle \L\frac{t^4}{4}-\frac{t^2}{2}\); \(\displaystyle \L\frac{(1)^4}{4}-\frac{(1)^2}{2}\); \(\displaystyle \L\frac{1}{4}-\frac{1}{2}\); \(\displaystyle \L\frac{1}{4}-\frac{2}{4} = -\frac{1}{4}\)
. . .
\(\displaystyle \L\frac{t^4}{4}-\frac{t^2}{2}\); \(\displaystyle \L\frac{(0)^4}{4}-\frac{(0)^2}{2} = 0\)
. . .
\(\displaystyle \L\frac{-1}{4}-0 = \frac{-1}{4}\)
. . .
\(\displaystyle \L4-2=2\)
. . .
Ok, for my final answer, I got 1.75, which is close to the answers given, but not exact. So, I figure that I must have made a small error somewhere.
 
I'm thinking that since the path goes under the axis from 0 to 1, an above from 1 to 2, so you have to swich things around for the second integral.

And for your second step, you have already integrated so you no longer need the integration symbols.
 
Note: "distance travelled" usually means "entire distance wandered, including back-tracking"; this differs from "displacement" which usually means "the net change in position, so back-tracking doesn't count".

Example: "Three steps forward; two steps back." In "distance travelled", this is five steps. In "displacement", this is one step. It's that difference that can make life very frustrating sometimes. :wink:

Assuming these definitions to hold, you first need to take the velocity and figure out where it is positive (going forward) and where it is negative (going backward), because you're going to have to integrate these separately if you're going to get the entire distance covered, not just the net change in position.

Eliz.
 
try this : \(\displaystyle \int\limits_1^2 {(t^3 } - t) - \int\limits_0^1 {(t^3 } - t)dt\)
Positive - Negative
 
Remember from before, absolute value!
\(\displaystyle \L
\int_0^2 {\left| {t^3 - t} \right|dt} = \int_0^1 {\left( { - t^3 + t} \right)dt} + \int_1^2 {\left( {t^3 - t} \right)dt}\)
 
stapel said:
Assuming these definitions to hold, you first need to take the velocity and figure out where it is positive (going forward) and where it is negative (going backward), because you're going to have to integrate these separately if you're going to get the entire distance covered, not just the net change in position.
Click to expand...

Man, I thought that I did that... *ponders*
 
Lizzie said:
\(\displaystyle \L\frac{-1}{4}-0 = \frac{-1}{4}\)
. . .
\(\displaystyle \L4-2=2\)
Click to expand...
I believe that was the first and second part. So I did 2+(-1/4) = 1.75... is there an error in any of that...I mean, I know that there is an error, but where?
 
Be careful of your notation. After you've integrated, you shouldn't still have the "integral" symbol in front or the "dx" at the end. And since you're doing the positive part and the negative part separately, flipping the sign on the negative part (in other words, integrating |v(t)| rather than v(t) itself), you have to start with that integration. You' can't

. . .int [0,1] [-v(t)] dt + int [1,2] [v(t)] dt

. . . . .= int [0,1] [t - t<sup>3</sup>] dt + int [1,2] [t<sup>3</sup> - t] dt

. . . . .= [(1/2)t<sup>2</sup> - (1/4)t<sup>4</sup>][0,1] + [(1/4)t<sup>4</sup> - (1/2)t<sup>2</sup>][1,2]

I'm not getting "1.5". My guess is that you subtracted a bit that you should have added.

Eliz.
 
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