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### Differential calculus and motion in a straight line

If I drive 180 kilometres in two hours, then my average velocity is 90 kilometres per hour. However, my instantaneous velocity during the journey is displayed on the speedometer, and my velocity may range from 0 km/h to 110 km/h, if the latter is the speed limit. Neither my velocity nor my acceleration will be constant.

Just as an average velocity corresponds to the gradient of a chord on the position–time graph, so an instantaneous velocity corresponds to the gradient of a tangent. We know that, if the motion is described by a nicely behaved position function $$x(t)$$, then the function for the instantaneous velocity can be found as the derivative of $$x(t)$$ with respect to time. This is discussed in the module Introduction to differential calculus.

Similarly, if the velocity of the particle is described by a nicely behaved function $$v(t)$$, the function for the instantaneous acceleration can be found as the derivative of $$v(t)$$ with respect to time.

For motion in a straight line with constant acceleration, we have seen that the displacement of the particle can be determined from a velocity–time graph by finding the signed area between the graph and the $$t$$-axis. This leads us to a more general approach.

Integration is the inverse process of differentiation. Therefore, if the velocity function is known, the position function can be found by integration. Similarly, if the acceleration function is known, the velocity function can be found.

#### Instantaneous velocity and speed

From now on in this module, the words velocity and speed alone will mean instantaneous velocity and speed.

The instantaneous velocity $$v(t)$$ of a particle is the derivative of the position with respect to time. That is,

$v(t) = \dfrac{dx}{dt}.$

This derivative is often written as $$\dot{x}(t)$$, or simply as $$\dot{x}$$. From here on, the dot is used to denote the derivative with respect to $$t$$. This is Newton's notation.

#### Example

The position function of a particle is $$x(t) = 30t-5t^2$$.
1. Find the velocity function $$\dot{x}(t)$$.
2. Find the velocity when $$t=2$$.
3. Find the velocity when $$t=4$$.
4. When is the particle stationary?

#### Solution

1. if $$x(t)=30t-5t^2$$, then $$\dot{x}(t) = 30-10t$$.
2. $$\dot{x}(2) = 30-10 \times 2 = 10$$.
3. $$\dot{x}(4) = 30-10 \times 4 = -10$$.
4. $$\dot{x}(t) = 0$$ implies $$30-10t = 0$$, and so $$t = 3$$. The particle is stationary when $$t = 3$$.

#### Example

The position function of a particle is $$x(t) = 4\sin(2\pi t)$$.

1. Find the velocity function $$\dot{x}(t)$$.
2. Find the velocity when $$t=0$$.
3. Find the velocity when $$t=\dfrac{1}{4}$$.
4. Find the velocity when $$t=\dfrac{1}{2}$$.
5. Find the velocity when $$t=1$$.

#### Solution

1. If $$x(t) = 4\sin(2\pi t)$$, then $$\dot{x}(t) = 8\pi \cos(2\pi t)$$.
2. $$\dot{x}(0) = 8\pi$$.
3. $$\dot{x}(\dfrac{1}{4}) = 8\pi \cos \dfrac{\pi}{2} = 0$$.
4. $$\dot{x}(\dfrac{1}{2}) = 8\pi \cos \pi = -8\pi$$.
5. $$\dot{x}(1) = 8\pi \cos 2\pi = 8\pi$$.

A particle moving in a straight line is stationary when its velocity is zero, that is, $$\dfrac{dx}{dt} = 0$$. This is the origin of the term 'stationary point' introduced in the module Applications of differentiation.

When a stone is thrown into the air with a velocity of 30 m/s, the motion of the stone can be modelled by the position function $$x(t) = 30t-5t^2$$. The velocity function for this motion is $$\dot{x}(t) = 30-10t$$. The velocity is zero when $$t = 3$$. This occurs when the stone is at its maximum height.

The speed of a particle is the absolute value of its velocity. For example, if a particle is moving with a velocity of $$-5$$ m/s, then its speed is $$\bigl| -5 \bigr| = 5$$ m/s.

#### Acceleration

A particle whose velocity is changing is said to be accelerating. Acceleration is defined to be the rate of change of velocity. So the acceleration $$a(t)$$ is the derivative of the velocity with respect to time:

$a(t) = \dfrac{dv}{dt} = \dot{v}(t).$

The velocity is itself the derivative of the position, and so the acceleration is the second derivative of the displacement:

$a(t) = \dfrac{d^{2}x}{dt^2} = \ddot{x}(t).$

#### Example

A large stone is falling through a layer of mud. At time $$t$$ seconds, the depth of the stone in metres below the surface is given by $$x(t) = 20(1-e^{-\frac{t}{2}})$$.

1. Find $$\dot{x}(t)$$.
2. Find $$\ddot{x}(t)$$.
3. Find the position, velocity and acceleration when $$t = 1$$.
4. What happens to the position, velocity and acceleration as $$t\to \infty$$?

#### Solution

We take downwards to be the positive direction.

1. $$\dot{x}(t) = 10 e^{-\frac{t}{2}}$$.
2. $$\ddot{x}(t) = -5 e^{-\frac{t}{2}}$$.
3. $$x(1) = 20(1 - e^{-\frac{1}{2}}),\quad \dot{x}(1) = 10 e^{-\frac{1}{2}},\quad \ddot{x}(1) = -5 e^{-\frac{1}{2}}$$.

The direction of the velocity is down, as it has a positive sign, and the direction of the acceleration is up, as it has a negative sign. The velocity and the acceleration are in opposite directions, so the particle is slowing down.
4. As $$t\to \infty$$, we have $$x(t)\to 20$$, $$\dot{x}(t) \to 0$$ and $$\ddot{x}(t) \to 0$$.
##### Exercise 8

A particle moves in a straight line so that its position $$x(t)$$ metres at time $$t$$ seconds, relative to a fixed position $$O$$, is given by $$x(t) = t(t-4)^2$$. Find

1. the velocity at time $$t$$
2. the values of $$t$$ when the particle is instantaneously at rest
3. the acceleration after four seconds.

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