v(t) = t^2 - 4t m/s from t = 0 to t = 5. Find the displacement and… Step by Step

Step by step

The question

Question: find the displacement and the total distance for v(t) = t² − 4t from t = 0 to t = 5.

v(t) = t² − 4t

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 5 of (t² − 4t) dt

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

Now we have t³ over 3 − 2 t².

= (t³3 − 2t²) from 0 to 5

Put in 5, then take away the same with 0.

= (5³3 − 2(5)²) − (0³3 − 2(0)²)

That is −25/3 minus 0: −25/3.

= −253 − 0 = −253

So the displacement is −25/3 m.

displacement = −253 m

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

t² − 4t = 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

Draw v.

From 0 to 4, v is negative: it moves back.

From 4 to 5, v is positive: it moves forward.

3 The total distance

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

∫ from 0 to 4 of (t² − 4t) dt = −323

From 4 to 5: the integral is 7/3, the same way as before.

∫ from 4 to 5 of (t² − 4t) dt = 73

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 = −253 ✗

Distance counts every piece as positive.

Add them: 13 m.

distance = |−323| + |73| = 13 m

4 Check

Let's check: the pieces with their signs add up to −25/3, the displacement.

Right!

−323 + 73 = −253

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Answer