v(t) = 6 - 2t ft/s from t = 1 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) = 6 − 2t from t = 1 to t = 5.

v(t) = 6 − 2t

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 1 to 5 of (6 − 2t) dt

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

Now we have 6t − t².

= (6t − t²) from 1 to 5

Put in 5, then take away the same with 1.

= (6(5) − 5²) − (6(1) − 1²)

That is 5 minus 5: 0.

= 5 − 5 = 0

So the displacement is 0 ft.

displacement = 0 ft

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

6 − 2t = 0

Factor it.

−2(t − 3) = 0

v is 0 at t = 3.

t = 3

Draw v.

From 1 to 3, v is positive: it moves forward.

From 3 to 5, v is negative: it moves back.

3 The total distance

From 1 to 3: the integral is 4, the same way as before.

∫ from 1 to 3 of (6 − 2t) dt = 4

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

∫ from 3 to 5 of (6 − 2t) dt = −4

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: 8 ft.

distance = |4| + |−4| = 8 ft

4 Check

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

Right!

4 − 4 = 0

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Answer