Find the displacement and the total distance the particle travels… Step by Step

Step by step

The question

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

v(t) = t² − 5t + 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: in a tiny time it moves about v times that time.

Going back counts as negative.

displacement = ∫ from 0 to 5 of (t² − 5t + 4) dt

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

Now we have t³ over 3 − 5t² over 2 + 4t.

= (t³3 − 5t²2 + 4t) from 0 to 5

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

= (5³3 − 5(5)²2 + 4(5)) − (0³3 − 5(0)²2 + 4(0))

That is −5/6 minus 0: −5/6.

= −56 − 0 = −56

So the displacement is −5/6 m.

displacement = −56 m

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

t² − 5t + 4 = 0

To factor t² − 5t + 4, the rule: we need two numbers that multiply to 4 and add to −5.

They are −1 and −4: they multiply to 4 and add to −5.

? × ? = 4

? + ? = −5

−1 × (−4) = 4, −1 + (−4) = −5 ✓

So we write it like this.

(t − 1)(t − 4) = 0

v is 0 at t = 1 and t = 4.

t = 1 or 4

Draw v.

From 0 to 1, v is positive: it moves forward.

From 1 to 4, v is negative: it moves back.

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

3 The total distance

Call the antiderivative F.

Work it out at 1 and 4 too.

F(1) = 1³3 − 5(1)²2 + 4(1)

= 13 − 52 + 4

= 116

F(4) = 4³3 − 5(4)²2 + 4(4)

= 643 − 40 + 16

= −83

From 0 to 1: the same antiderivative at 1, minus at 0.

11/6 minus 0 is 11/6.

∫ from 0 to 1 of (t² − 5t + 4) dt = 116 − 0 = 116

From 1 to 4: the same antiderivative at 4, minus at 1.

−8/3 minus 11/6 is −9/2.

∫ from 1 to 4 of (t² − 5t + 4) dt = −83 − 116 = −92

From 4 to 5: the same antiderivative at 5, minus at 4.

−5/6 minus −8/3 is 11/6.

∫ from 4 to 5 of (t² − 5t + 4) dt = −56 − (−83) = 116

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 = −56 ✗

Distance counts every piece as positive.

Add them: 49/6 m.

distance = |116| + |−92| + |116| = 496 m

4 Check

Let's check: the pieces with their signs add up to −5/6, the displacement.

Right!

116 − 92 + 116 = −56

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Answer