v(t) = 3t^2 - 12 m/s from t = 0 to t = 3. Find the displacement and… Step by Step

Step by step

The question

Question: find the displacement and the total distance for v(t) = 3t² − 12 from t = 0 to t = 3.

v(t) = 3t² − 12

Displacement is where it ends up, compared with the start.

Distance counts every step, forward and back.

displacement: where it ends up

distance: how far it travels

1 The net change

The displacement is the integral of v.

Going back counts as negative.

displacement = ∫ from 0 to 3 of (3t² − 12) dt

The antiderivative: add 1 to each power, divide by the new power.

Now we have t³ − 12t.

= (t³ − 12t) from 0 to 3

Put in 3, then take away the same with 0.

= (3³ − 12(3)) − (0³ − 12(0))

That is −9 minus 0: −9.

= −9 − 0 = −9

So the displacement is −9 m.

displacement = −9 m

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

3t² − 12 = 0

x² − 4 is a difference of two squares: x² minus 2 squared.

So it is (x − 2)(x + 2).

3(t + 2)(t − 2) = 0

v is 0 at t = 2.

t = 2

3 The total distance

From 0 to 2: the integral is −16, the same way as before.

∫ from 0 to 2 of (3t² − 12) dt = −16

From 2 to 3: the integral is 7, the same way as before.

∫ from 2 to 3 of (3t² − 12) dt = 7

Careful!

The integral of v is the displacement.

Going back cancels going forward.

For the distance, add each piece without its sign.

Careful: not distance = −9 ✗

Distance counts every piece as positive.

Add them: 23 m.

distance = |−16| + |7| = 23 m

4 Check

Let's check: the pieces with their signs add up to −9, the displacement.

Right!

−16 + 7 = −9

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Answer