15.5 Calculus in kinematics — Learn

Prerequisites: differentiating polynomials; integrating polynomials; finding a constant of integration from an initial condition; the constant-acceleration formulae from Section 15.4.

By the end of this section you can: find a velocity by differentiating a displacement; find an acceleration by differentiating a velocity; recover a velocity and a displacement by integrating an acceleration; use $v = 0$ to find when an object is at rest; tell a displacement apart from a distance when the object turns around.

1. Differentiating displacement

Suppose the displacement of a particle along a straight line is a function of time,

The velocity is the rate of change of displacement, so

The acceleration is the rate of change of velocity:

The unit of $v$ is m/s and the unit of $a$ is m/s$^2$, because $t$ is in seconds and $s$ is in metres.

Example 1. The displacement of a particle is $s = t^3 - 6t^2 + 9t$ metres at time $t$ seconds. Find its velocity and its acceleration.

Differentiate once for the velocity:

Differentiate again for the acceleration:

2. Going backwards with integration

If you know the acceleration, integrate to find the velocity. If you know the velocity, integrate to find the displacement.

Each integration brings a constant of integration. The initial conditions, such as a starting velocity or a starting position, fix the constants.

Example 2. A particle moves with acceleration $a = 6t$ m/s$^2$. Its velocity is $4$ m/s when $t = 0$, and its displacement is $0$ m when $t = 0$. Find the velocity and the displacement.

Integrate the acceleration:

At $t = 0$, $v = 4$, so $C = 4$ and

Now integrate the velocity:

At $t = 0$, $s = 0$, so $K = 0$ and

3. Constant acceleration from calculus

For a particle with constant acceleration $a$, integrating once gives

where $u$ is the velocity at $t = 0$. Integrating again gives

where the displacement is measured from the position at $t = 0$. So the calculus method and the constant-acceleration formulae are two views of the same motion. The calculus method works even when the acceleration is not constant.

4. At rest, and turning around

The particle is at rest when $v = 0$. Solving this equation gives the times when the particle stops. If the sign of the velocity changes, the particle turns around at that time.

To find a distance travelled, split the time interval at each moment when $v = 0$. Add the sizes of the displacements on the separate parts. The displacement over the whole interval is the sum with signs.

Example 3. A particle has displacement $s = 4t - t^2$ metres. Find when it is at rest, and find its displacement and distance travelled from $t = 0$ to $t = 5$ s.

The velocity is

Set $v = 0$: $4 - 2t = 0$, so the particle is at rest at $t = 2$ s.

At $t = 0$, $s = 0$. At $t = 2$, $s = 8 - 4 = 4$ m. At $t = 5$, $s = 20 - 25 = -5$ m.

The displacement over the whole interval is $s(5) - s(0) = -5$ m. The particle moves forward $4$ m and then backward $9$ m, so the distance travelled is

Progress check

Return to the free kick. For the vertical motion, take upwards as positive and let $y$ be the height in metres after $t$ seconds. The acceleration is constant, $a = -9.8$ m/s$^2$. The initial vertical velocity is $u = 10\sqrt{2}$ m/s, and the ball starts at $y = 0$.

1. Integrate the acceleration to find $v$ as a function of $t$.
2. Integrate the velocity to find $y$ as a function of $t$.
3. Substitute $t = 2\sqrt{2}$ and work out the height, using $10\sqrt{2} \times 2\sqrt{2} = 40$ and $(2\sqrt{2})^2 = 8$.
4. Compare the height with the crossbar at $2.44$ m.

Watch out. Three mistakes are common with this calculus.

A derivative used where an integral is needed. Velocity is the derivative of displacement, so going from displacement to velocity is a differentiation. Going from acceleration to velocity is the opposite move, an integration. Read which quantities you are given before you choose.

The constant of integration dropped. Without the initial condition, $\int 6t\,dt$ can be any of $3t^2 + C$. In Example 2 the condition $v = 4$ at $t = 0$ fixes $C = 4$ and gives $v = 3t^2 + 4$. A velocity left as $3t^2 + C$ is not an answer.

Displacement used as distance. When the particle turns around, the displacement for the whole interval can be smaller than the distance, or even negative. In Example 3 the displacement is $-5$ m and the distance is $13$ m. Re-check: find every time when $v = 0$, split the interval there, and add the sizes of the separate displacements to get the distance.

You should now be able to

• find a velocity by differentiating a displacement function;
• find an acceleration by differentiating a velocity function;
• find a velocity or a displacement by integrating and using an initial condition;
• derive $v = u + at$ and $s = ut + \frac{1}{2}at^2$ from a constant acceleration;
• find the times when a particle is at rest;
• split an interval at a turning point and find a total distance.