Calculating acceleration

Here, $u$ is the initial velocity and $v$ is the final velocity, both measured in metres per second (m/s). The symbol $\Delta$ means ‘change in’, so $\Delta v = v-u$. Time taken, $t$, is measured in seconds (s), and acceleration, $a$, in metres per second squared (m/s²).

The unit follows from the calculation: dividing a velocity change in m/s by a time in seconds gives m/s². You can think of this as ‘metres per second, per second’.

A family car taking about $10\text{ s}$ to increase its velocity from rest to $27\text{ m/s}$ has an average acceleration of:

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

Its velocity increases by an average of $2.7\text{ m/s}$ each second. This does not mean the increase must be exactly the same in every second: an average can describe motion whose acceleration varies.

The equation can also be rearranged to find a velocity change or a time:

$$\Delta v = at \qquad t = \frac{\Delta v}{a}$$

Slowing down and the sign of acceleration

An object that slows down is decelerating. If it is moving in the positive direction, its final velocity is smaller than its initial velocity, so its change in velocity and acceleration are negative.

For example, a train moving in the positive direction slows from $50\text{ m/s}$ to $42\text{ m/s}$ in $30\text{ s}$:

$$\Delta v = 42-50 = -8\text{ m/s}$$ $$a = \frac{-8}{30} \approx -0.27\text{ m/s}^2$$

The negative sign shows that its acceleration acts opposite to its positive motion. The magnitude of its deceleration is $0.27\text{ m/s}^2$.

A negative acceleration does not always mean slowing down: its meaning depends on the direction of motion. Deceleration means acceleration opposite to the object's motion.

Estimating everyday accelerations

To estimate an acceleration, estimate how much velocity changes and how long that change takes, then divide. A greater velocity change in the same time gives a greater acceleration; the same change spread over a longer time gives a smaller acceleration.

The family-car calculation provides a useful reference: an average acceleration of roughly 2–3 m/s² is plausible for a car accelerating from rest. For comparison, an object falling freely near the Earth's surface accelerates at about 10 m/s². These are useful scales for judging an estimate, not fixed accelerations for every car or every falling object.

Uniform acceleration when time is not known

Uniform acceleration means constant acceleration: velocity changes by equal amounts in equal time intervals. For straight-line motion with uniform acceleration, another equation connects velocity, acceleration and distance:

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

Here, $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $s$ is distance travelled in metres (m). The velocities are in m/s and acceleration is in m/s². This equation is especially useful when time is not given or required.

Consider a car accelerating steadily from rest at $2.5\text{ m/s}^2$ until it reaches $16\text{ m/s}$. ‘From rest’ means $u=0\text{ m/s}$, and ‘steadily’ means the acceleration is constant. Rearranging for distance gives:

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

Substituting:

$$s = \frac{16^2-0^2}{2\times2.5} = \frac{256}{5} = 51.2\text{ m}$$

The car travels $51.2\text{ m}$ while gaining this velocity. If acceleration were required instead, the rearrangement would be $a = \frac{v^2-u^2}{2s}$. Choose the equation by looking at the quantities available and checking whether acceleration is constant.

• $u$: initial velocity; $v$: final velocity; $\Delta v$: change in velocity — all in m/s.
• $t$: time taken in s; $a$: acceleration in m/s².
• Rearrangements: $\Delta v = at$ and $t = \frac{\Delta v}{a}$.
• Constant velocity means zero acceleration. Average acceleration need not be constant throughout the interval.

Deceleration and estimates

• Slowing down means decelerating: acceleration acts opposite to motion.
• For motion in the positive direction, slowing down gives negative acceleration.
• Magnitude is the size without the sign.
• Useful scales: family car 2–3 m/s²; free fall near Earth about 10 m/s².
• Estimate acceleration by dividing an estimated velocity change by the time taken.

Uniform acceleration

Use only for constant acceleration:

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

$s$ is distance in m. This equation does not require time. Starting from rest means $u=0$.