SQA Advanced Higher Physics

Rates of change

Kinematics describes motion without asking what causes it. At Advanced Higher the quantities are treated as functions of time, and the language of calculus is used because velocity and acceleration are rates of change.

Differential notation writes a rate of change as a derivative. If a quantity $y$ depends on time $t$, then $\dfrac{dy}{dt}$ is how fast $y$ is changing at that instant. It is the gradient of the graph of $y$ against $t$, taken at a single point, which is why it can be found for a curve as well as for a straight line.

For straight-line motion, with displacement $s$ measured from a fixed point:

Velocity measures how quickly displacement changes as time passes, and acceleration how quickly velocity changes, so acceleration is the second derivative of displacement. $v$ is in $\mathrm{m\,s^{-1}}$ and $a$ in $\mathrm{m\,s^{-2}}$. Because displacement, velocity and acceleration are vectors, in one dimension their sign carries the direction: a negative $v$ means motion in the negative direction, and a negative $a$ with a positive $v$ means the object is slowing down.

Deriving the equations of motion

The equations for constant acceleration follow from the definitions by integration, so they do not need to be learned as separate facts. They apply only when $a$ is constant.

Start from $a = \dfrac{dv}{dt}$ with $a$ constant, and let the velocity be $u$ at $t = 0$ and $v$ at time $t$. Integrating both sides with respect to time between these limits gives

Now use $v = \dfrac{ds}{dt} = u + at$, and integrate again from displacement $0$ at $t = 0$ to displacement $s$ at time $t$:

For the third equation, write the acceleration as $a = \dfrac{dv}{dt} = \dfrac{dv}{ds}\dfrac{ds}{dt} = v\dfrac{dv}{ds}$. Then $a\,ds = v\,dv$, and for constant $a$

The three equations are therefore

where $u$ is the initial velocity, $v$ the final velocity, $s$ the displacement, $a$ the constant acceleration and $t$ the time.

Varying acceleration

When the acceleration is not constant the three equations cannot be used. The displacement has to be given as a function of time, and velocity and acceleration are found by differentiating it. If the acceleration is given as a function of time, velocity and displacement are found by integrating, and each integration introduces a constant that the starting conditions fix.

Worked example. A trolley's displacement from a starting point is $s = 3t^2 - 0.5t^3$, with $s$ in metres and $t$ in seconds.

At $t = 2.0\,\mathrm{s}$, $s = 12 - 4 = 8.0\,\mathrm{m}$, $v = 12 - 6.0 = 6.0\,\mathrm{m\,s^{-1}}$ and $a = 6 - 6 = 0$. The acceleration is zero there, which is where the velocity is a maximum, because $v$ stops increasing when $\dfrac{dv}{dt} = 0$. At $t = 4.0\,\mathrm{s}$ the velocity is $24 - 24 = 0$, so the trolley is momentarily at rest, and $a = 6 - 12 = -6.0\,\mathrm{m\,s^{-2}}$ shows it is about to move back.

Going the other way, suppose an object starts from rest with acceleration $a = 4t$ in $\mathrm{m\,s^{-2}}$. Then

Starting from rest gives $v = 0$ at $t = 0$, so $c = 0$. If it also starts at the origin then $c' = 0$. After $3.0\,\mathrm{s}$ the speed is $18\,\mathrm{m\,s^{-1}}$ and the displacement $18\,\mathrm{m}$.

Gradients on motion graphs

The gradient of a line on a graph is the rate of change of the quantity on the vertical axis with respect to the quantity on the horizontal axis. For a curve the gradient at a point is the gradient of the tangent there, which is the derivative at that point.

A straight line on a $v$-$t$ graph has a constant gradient, so constant acceleration. A curved line shows a changing acceleration, and the acceleration at a chosen moment is found by drawing a tangent at that time and calculating its gradient, using two points as far apart on the tangent as the graph allows.

Areas under graphs

Integration finds the area between a curve and the horizontal axis. The area under a velocity against time graph between times $t_1$ and $t_2$ is

and the area under an acceleration against time graph between the same limits is the change in velocity,

Area below the time axis counts as negative, which reflects movement in the opposite direction. The total displacement is the signed sum, and the distance travelled is the sum of the areas without their signs.

When the line is straight, the area is a rectangle, triangle or trapezium and can be found without integrating. Worked example: an acceleration increases uniformly from $0$ to $6.0\,\mathrm{m\,s^{-2}}$ over $4.0\,\mathrm{s}$. The area under the graph is a triangle, $\tfrac12 \times 4.0 \times 6.0 = 12\,\mathrm{m\,s^{-1}}$, which is the change in velocity. If the graph is a curve given by an expression, the same result comes from the integral: with $a = 1.5t$, $\Delta v = \int_0^4 1.5t\,dt = \left[0.75t^2\right]_0^4 = 12\,\mathrm{m\,s^{-1}}$.

Where a graph is a smooth curve with no equation, the area is estimated by counting squares or by splitting it into strips, and the gradient by drawing a tangent. The estimate has an uncertainty from the reading of the graph.

Choosing a method

A problem with constant acceleration and known quantities uses the three equations. A problem with an expression for $s$, $v$ or $a$ in terms of $t$ uses differentiation to move towards acceleration and integration to move towards displacement. A problem that gives a graph uses gradients and areas, and a graph that is straight-sided can be handled with geometry.