Expert: Paul Klarreich - 1/26/2010

Question
I asked you this earlier and got the answer for total displacement but now when I try and find distance I got the same as the total displacement which was (-20/3). I tried both (-20/3) and (20/3) and they were both wrong. how do I find the distance to this problem:

The velocity function is -t^2+4t-3 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [0,5]

For the Total Displacement I got (-20/3) which is correct.  

Answer
Questioner: brooke
Country: United States
Category: Calculus
Private: No
Subject: velocity problem
Question: I asked you this earlier and got the answer for total displacement but now when I try and find distance I got the same as the total displacement which was (-20/3). I tried both (-20/3) and (20/3) and they were both wrong. how do I find the distance to this problem:

The velocity function is -t^2+4t-3 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [0,5]

For the Total Displacement I got (-20/3) which is correct.
.............................................................
There are many rules for success in mathematics.  The first ten say:

Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.
Make sure you know the vocabulary terms correctly.


If  velocity is v(t), then:

Displacement =

{t=b
|    v(t) dt
}t=a

Distance traveled =

{t=b
|   abs(v(t)) dt
}t=a


Notice the difference?  If you drive from Miami to New York and back, your displacement is zero, but your odometer still says you traveled 2600 miles.

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So if:

The velocity function is -t^2+4t-3 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [0,5]

Here is what you must do:

1. Find where  -t^2+4t-3 is positive.  In that Interval(or intervals), integrate -t^2+4t-3 itself, because it is positive.

2. Find where  -t^2+4t-3 is negative.  In that Interval(or intervals), integrate -(-t^2+4t-3), which will be positive.

3. You will get two (or more) positive distances.  Add them all up.

Hint: y = -t^2+4t-3  is a parabola opening down, and crossing the x-axis (t-axis?) at  t = 1, t = 3.

So you will do three integrals:

{t=1
|   either (-t^2+4t-3) dt or -(-t^2+4t-3) dt
}t=0

{t=3
|   either (-t^2+4t-3) dt or -(-t^2+4t-3) dt
}t=1

{t=5
|   either (-t^2+4t-3) dt or -(-t^2+4t-3) dt
}t=3

I leave it to you to do these three integrals.  (Another hint:  If you do one and get a negative answer, you blew it.)