Motion with integrals: position, displacement, distance

About this lesson

Calculus BC (AP-aligned) › Applications of integration

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Motion with integrals: position, displacement, distance

Warm-up

Three quick questions from earlier lessons.

Discover

A particle moves right for 2 s at 3 m/s, then left for 1 s at 3 m/s. Its displacement is 6−3=36 - 3 = 3 m. How far did it travel in total?

  1. 9 m
  2. 3 m

Key idea

Integrals of velocity give position changes; integrals of speed give distance.

s(b)=s(a)+∫abv(t) dts(b) = s(a) + \displaystyle\int_a^b v(t)\,dt. Displacement: ∫abv(t) dt\displaystyle\int_a^b v(t)\,dt. Total distance: ∫ab∣v(t)∣ dt\displaystyle\int_a^b |v(t)|\,dt, found by splitting where vv changes sign.

Worked example

v(t)=t−2v(t) = t - 2 on [0,5][0, 5]. Displacement and distance.

Displacement =12.5−10=2.5= 12.5 - 10 = 2.5
∫(t − 2) dt = t²/2 − 2t from 0 to 5.
∫02∣v∣=2\displaystyle\int_0^2 |v| = 2, ∫25∣v∣=4.5\displaystyle\int_2^5 |v| = 4.5
v < 0 on [0, 2]: that piece has area 2.
Distance =6.5= 6.5
Add the pieces.

Your turn, guided

Your turn: v(t)=3t2v(t) = 3t^2, s(0)=4s(0) = 4. Find s(2)s(2).

s(2)=12s(2) = 12
4 + 8.

Practice ★

  1. A particle has velocity v(t)=2t+7v(t) = 2t + 7 and starts at s(0)=4s(0) = 4. Find its position at t=5t = 5.
  2. A particle has velocity v(t)=2t+2v(t) = 2t + 2 and starts at s(0)=8s(0) = 8. Find its position at t=5t = 5.
  3. A particle has velocity v(t)=2t+9v(t) = 2t + 9 and starts at s(0)=1s(0) = 1. Find its position at t=5t = 5.

Watch out!

Practice ★★

  1. A particle's velocity is v(t)=t−4v(t) = t - 4 m/s. Find the total distance it travels from t=0t = 0 to t=6t = 6, in meters.
  2. A particle's velocity is v(t)=t−4v(t) = t - 4 m/s. Find the total distance it travels from t=0t = 0 to t=9t = 9, in meters.
  1. v(t)v(t) is a velocity in m/s. What does ∫09∣v(t)∣ dt\displaystyle\int_0^{ 9 } |v(t)|\,dt give?
    1. The total distance traveled from t=0t = 0 to t=9t = 9
    2. The displacement from t=0t = 0 to t=9t = 9
    3. The average speed on [0,9][0, 9]

Exit ticket

Three questions to check you have got it.

  1. A particle has velocity v(t)=2t+7v(t) = 2t + 7 and starts at s(0)=4s(0) = 4. Find its position at t=5t = 5.
  2. A particle has velocity v(t)=2t+2v(t) = 2t + 2 and starts at s(0)=8s(0) = 8. Find its position at t=5t = 5.
  3. A particle has velocity v(t)=2t+9v(t) = 2t + 9 and starts at s(0)=1s(0) = 1. Find its position at t=5t = 5.

Summary card

Motion with integrals

  • Position: start + ∫v\int v.
  • Displacement ∫v\int v; distance ∫∣v∣\int |v|.

Watch out: split where vv changes sign.

End of the lesson

Next lesson: Area between curves Back to the chapter

Answers: Motion with integrals: position, displacement, distance

A

  1. 6464
  2. 4343
  3. 7171

B

  1. 1010
  2. 412\frac{41}{2}

C

  1. The total distance traveled from t=0t = 0 to t=9t = 9

D

  1. 6464
  2. 4343
  3. 7171

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