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

Step by step

The question

Question: find the displacement and the total distance for v(t) = 2t − 4 from t = 0 to t = 5.

v(t) = 2t − 4

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 (2t − 4) dt

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

Now we have t² − 4t.

= (t² − 4t) from 0 to 5

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

= (5² − 4(5)) − (0² − 4(0))

That is 5 minus 0: 5.

= 5 − 0 = 5

So the displacement is 5 m.

displacement = 5 m

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

2t − 4 = 0

Factor it.

2(t − 2) = 0

v is 0 at t = 2.

t = 2

Draw v.

From 0 to 2, v is negative: it moves back.

From 2 to 5, v is positive: it moves forward.

3 The total distance

From 0 to 2: the integral is −4, the same way as before.

∫ from 0 to 2 of (2t − 4) dt = −4

From 2 to 5: the integral is 9, the same way as before.

∫ from 2 to 5 of (2t − 4) dt = 9

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 = 5 ✗

Distance counts every piece as positive.

Add them: 13 m.

distance = |−4| + |9| = 13 m

4 Check

Let's check: the pieces with their signs add up to 5, the displacement.

Right!

−4 + 9 = 5

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Answer