v(t) = 4t - t^2 m/s from t = 0 to t = 6. Find the displacement and… Step by Step

Step by step

The question

Question: find the displacement and the total distance for v(t) = 4t − t² from t = 0 to t = 6.

v(t) = 4t − t²

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 6 of (4t − t²) dt

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

Now we have 2 t² − t³ over 3.

= (2t² − t³3) from 0 to 6

Put in 6, then take away the same with 0.

= (2(6)² − 6³3) − (2(0)² − 0³3)

That is 0 minus 0: 0.

= 0 − 0 = 0

So the displacement is 0 m.

displacement = 0 m

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

4t − t² = 0

We need two numbers that multiply to 0 and add to −4: 0 and −4.

−t(t − 4) = 0

v is 0 at t = 4.

t = 4

3 The total distance

From 0 to 4: the integral is 32/3, the same way as before.

∫ from 0 to 4 of (4t − t²) dt = 323

From 4 to 6: the integral is −32/3, the same way as before.

∫ from 4 to 6 of (4t − t²) dt = −323

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 = 0 ✗

Distance counts every piece as positive.

Add them: 64/3 m.

distance = |323| + |−323| = 643 m

4 Check

Let's check: the pieces with their signs add up to 0, the displacement.

Right!

323 − 323 = 0

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Answer