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Calculus Applications of integration

Motion problems (position, velocity, acceleration)

20 practice questions 0 video lessons Theory + worked examples

Motion Problems

California Calculus • Standard 16.0 • Applications of Integration

Motion Problems is a topic in Applications of Integration in the California Calculus Standards. It is aligned to Standard 16.0, which requires students to use definite integrals in problems involving velocity and acceleration, interpreting the derivative as a rate of change (Standard 4.2).

Motion problems connect position, velocity, and acceleration through calculus: velocity is \(s'(t)\) and acceleration is \(v'(t)\), while displacement is \(\int v\,dt\) and total distance is \(\int|v|\,dt\).

California Calculus › Applications of Integration › Motion Problems  —  Standard 16.0

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Theory

In motion problems, velocity is the derivative of position and acceleration the derivative of velocity; integration reverses each step. Displacement is \(\int v\,dt\) (signed), and total distance is \(\int |v|\,dt\). Units are feet and ft/s.

Position \(s(t)\), velocity \(v(t)\), and acceleration \(a(t)\) are linked by calculus:

\[v=\dfrac{ds}{dt},\qquad a=\dfrac{dv}{dt}.\]

Going backward, integrate: \(v=\int a\,dt\) and \(s=\int v\,dt\), each with a constant fixed by an initial condition.

Displacement over \([a,b]\) is the signed integral \(\int_a^b v\,dt\); total distance uses \(|v|\), splitting where \(v\) changes sign.
Key idea: displacement can be zero while distance is large — the object goes out and comes back.
A velocity-time graph crossing zero On a velocity-time graph, area above the axis is forward displacement and area below is backward; the object reverses where velocity crosses zero. t v − + v(t)
A velocity-time graph: area above is forward, below is backward.
Position and velocity against time Velocity is the derivative of position; position is the integral of velocity, recovered with an initial condition. t s(t) v = s′
Position \(s(t)\) and velocity \(v=s'\).

Derivatives one way, integrals the other:

\[v=\dfrac{ds}{dt},\quad a=\dfrac{dv}{dt};\qquad s=\int v\,dt,\quad v=\int a\,dt\]
velocity and acceleration by differentiation; position and velocity by integration
\[\text{displacement}=\int_a^b v\,dt,\qquad\text{distance}=\int_a^b |v|\,dt\]
displacement is the integral of velocity; distance is the integral of its absolute value
Signs matter. A negative velocity is motion backward; displacement keeps the sign, distance does not.

How to solve a motion problem

  1. Differentiate to go position \(\to\) velocity \(\to\) acceleration.
  2. Integrate to go back, using initial conditions to fix each constant.
  3. Displacement is \(\int v\,dt\); for total distance integrate \(|v|\), splitting where \(v=0\).
Example 1 — Displacement from velocity
A particle has \(v(t)=3t^2\) ft/s. Find its displacement on \([0,2]\).
Solution

Displacement \(=\displaystyle\int v\,dt\); antiderivative \(t^3\).

\(\int_0^2 3t^2\,dt\)\(=\)\(\big[t^3\big]_0^2\)
\(=\)\(2^3-0^3=8\)

Displacement \(=8\) ft.

displacement equals 8 feet
Example 2 — Velocity from position
For \(s(t)=t^3\) ft, find the velocity at \(t=4\) s.
Solution

Velocity is \(s'(t)\).

\(v(t)\)\(=\)\(3t^2\)
\(v(4)\)\(=\)\(48\)

The velocity is \(48\) ft/s.

velocity equals 48 feet per second
Example 3 — Total distance
A particle has \(v(t)=t-2\) ft/s on \([0,4]\). Find the total distance.
Solution

Total distance \(=\displaystyle\int_0^4 |v|\,dt\); \(v=t-2\) changes sign at \(t=2\), so split there and integrate each piece.

\(\int_0^2 (2-t)\,dt\)\(=\)\(\left[2t-\dfrac{t^2}{2}\right]_0^2=2\)
\(\int_2^4 (t-2)\,dt\)\(=\)\(\left[\dfrac{t^2}{2}-2t\right]_2^4=2\)
\(\text{total}\)\(=\)\(2+2=4\)

Total distance \(=4\) ft (displacement is \(0\)).

total distance equals 4 feet
Example 4 — Velocity from acceleration
An object has \(a=4\) ft/s\(^2\) and \(v(0)=2\) ft/s. Find \(v(t)\).
Solution

Integrate acceleration; the constant is the initial velocity.

\(v(t)\)\(=\)\(\int 4\,dt=4t+C\)
\(v(0)=2\)\(\Rightarrow\)\(v(t)=4t+2\)
velocity equals 4t plus 2

Common pitfalls

Displacement vs distance. Displacement is signed; total distance integrates \(|v|\) and is never negative.
Do not forget the constant. Recovering \(v\) or \(s\) by integration needs an initial condition to pin down \(C\).
Acceleration is the second derivative. \(a=s''\); a constant acceleration integrates to a linear velocity.

Frequently asked questions

What is the difference between displacement and distance?

Displacement is the signed change in position, \(\int v\,dt\); total distance integrates \(|v|\) and counts all travel as positive.

How do you find velocity from position?

Differentiate: \(v(t)=s'(t)\). To go the other way, integrate the velocity to recover position.

How do you find position from velocity?

Integrate: \(s(t)=\int v\,dt\), then use an initial position to determine the constant of integration.

How do you find total distance traveled?

Integrate the absolute value of velocity, \(\int |v|\,dt\), splitting the integral wherever \(v\) changes sign.

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