Step by step
The question
Question: find the displacement and the total distance for v(t) = 3 − t from t = 0 to t = 6.
Displacement is where it ends up, compared with the start.
Distance counts every step, forward and back.
1 The net change
The displacement is the integral of v.
Going back counts as negative.
The antiderivative: add 1 to each power, divide by the new power.
Now we have 3t − t² over 2.
Put in 6, then take away the same with 0.
That is 0 minus 0: 0.
So the displacement is 0 m.
2 Where it turns around
It turns around where v changes sign.
First, where is v = 0?
Factor it.
v is 0 at t = 3.
Draw v.
From 0 to 3, v is positive: it moves forward.
From 3 to 6, v is negative: it moves back.
3 The total distance
From 0 to 3: the integral is 9/2, the same way as before.
From 3 to 6: the integral is −9/2, the same way as before.
Careful!
The integral of v is the displacement.
Going back cancels going forward.
For the distance, add each piece without its sign.
Distance counts every piece as positive.
Add them: 9 m.
4 Check
Let's check: the pieces with their signs add up to 0, the displacement.
Right!
Let's look back.
First: the net change.
Then: where it turns around.
Then: the total distance.
Last: check.