Motion Equations and Everyday Accelerations

Acceleration describes how quickly motion changes

Velocity is speed in a specified direction. Acceleration is the rate at which velocity changes. For motion along a straight line, we can calculate average acceleration from the change in velocity and the time taken:

$$a = \frac{v-u}{t}$$

Here, $u$ is the initial velocity, $v$ is the final velocity and $t$ is the time interval. Velocities are measured in m/s, time in seconds and acceleration in m/s².

An acceleration of $3\text{ m/s}^2$ means that velocity changes by $3\text{ m/s}$ each second. If this change is the same every second, the acceleration is uniform. For example, a car accelerating uniformly from rest to $12\text{ m/s}$ in $4\text{ s}$ has an acceleration of $$.

Linking speed and distance without knowing time

Sometimes a question gives the distance over which an object accelerates, rather than the time taken. For uniform acceleration in a straight line, use:

$$v^2-u^2=2as$$

In this equation, $u$ and $v$ are the initial and final speeds in m/s, $a$ is the acceleration in m/s², and is the distance travelled in metres. It links the to acceleration and distance. It is useful because it contains no time variable.

Consider a car starting from rest and accelerating uniformly at $3\text{ m/s}^2$ over $24\text{ m}$. Starting from rest means $u=0$, so:

$$v^2 = u^2+2as = 0^2+2\times3\times24 = 144$$

Taking the square root gives a final speed of $12\text{ m/s}$. This is the speed at the end of the distance, not the average speed over the journey.

Selecting and rearranging the equation

Start by identifying the initial speed, final speed, acceleration and distance. If three are known and the fourth is required, the relationship can be rearranged to find it:

$$v=\sqrt{u^2+2as}$$ $$a=\frac{v^2-u^2}{2s}$$ $$s=\frac{v^2-u^2}{2a}$$

The equation also describes uniform slowing down. Taking the direction of travel as positive, the acceleration is negative: the final speed is smaller than the initial speed, so $v^2-u^2$ is negative too. If the object stops, its final speed is zero. The distance calculated is the distance covered while slowing down.

Estimating everyday accelerations

An estimate uses realistic approximate quantities rather than exact measurements. To estimate an acceleration, choose a sensible initial speed, final speed and time interval, then divide the change in speed by the time taken.

A typical family car might increase its speed from $0\text{ m/s}$ to about $27\text{ m/s}$ in roughly $10\text{ s}$. Its average acceleration is therefore about:

$$a=\frac{27-0}{10}=2.7\text{ m/s}^2$$

This gives a useful scale: a family car's acceleration is typically a few m/s². It does not mean that the car's acceleration stays exactly constant throughout those ten seconds. The calculation gives an average over the interval.

The magnitude of an acceleration is its size, without a direction or sign. When estimating how strongly something slows down, divide the size of its speed change by the time taken and report a positive magnitude.

Free fall near Earth's surface

An object is in free fall when gravity is the only force acting on it. With air resistance absent or negligible, all objects near Earth's surface have the same downward acceleration, regardless of their mass:

$$g\approx9.8\text{ m/s}^2$$

For an object dropped from rest, this means its downward speed increases by approximately $9.8\text{ m/s}$ each second. This acceleration is much larger than the typical acceleration of a family car.

The condition about air resistance matters. Objects falling through air do not necessarily accelerate at $9.8\text{ m/s}^2$, because air resistance can affect their motion. Use the free-fall value when gravity alone determines the acceleration.

Uniform-acceleration equation

$$v^2-u^2=2as$$

• $u$: initial speed; : — both in m/s.