1.10 - Constant acceleration equation

The relationship and its symbols

The Edexcel relationship for this lesson is:

Here:

• $u$ is the initial speed, in m/s
• $v$ is the final speed, in m/s
• $a$ is the acceleration, in $\mathrm{m/s^2}$
• $\mathrm{s}$ is the distance moved, in m

Here motion is along a straight line without reversal, with the initial direction taken as positive. The equation connects two speeds with the distance covered during the same interval. More generally, u and v are signed velocities and s is displacement; do not substitute total path length for displacement if motion reverses. It is called a constant acceleration equation because it only works when the acceleration is constant over that interval.

Notice that the equation uses $v^2$ and $u^2$, not just $v$ and $u$. If you are finding a speed, your calculation will usually find $v^2$ or $u^2$ first; the final step is to take the square root.

When to use it

Use this equation when a question links:

• initial speed
• final speed
• acceleration
• distance moved

It is most useful when the time is not given and is not being asked for. If the question gives speed, acceleration and distance data, this equation often avoids needing to find the time first.

[DIAGRAM: velocity_time_constant_acceleration: Constant acceleration equation - diagram 01; asset_slug: p04_constant_acceleration_equation__velocity_time_area; recommended_method: matplotlib; description: NovaLearn pale-palette speed-time graph for constant acceleration, with u and v aligned to their levels, sage area labelled distance moved s, a time-t bracket and a clay gradient triangle labelled acceleration a.]

The graph is not needed for every calculation, but it gives a useful picture. For constant acceleration, the velocity-time graph is a straight line. The slope represents acceleration, and the area under the line represents distance moved, so it makes sense that the final relationship connects $a$, $\mathrm{s}$, $u$ and $v$.

You can see where the equation comes from using the graph. The area under the straight line is the average speed multiplied by the time: $s = \frac{u + v}{2} \times t$. The slope gives the time: $t = \frac{v - u}{a}$. Put that time into the area and it cancels out: $s = \frac{(v + u)(v - u)}{2a} = \frac{v^2 - u^2}{2a}$. Multiply both sides by $2a$ and rearrange to get $v^2 = u^2 + 2as$. That is why the equation has no $t$ in it.

If an object is slowing down, the acceleration may be negative. For example, a car moving forwards and braking can have $a = -4.0 \mathrm{m/s^2}$. With forward velocity taken as positive, negative acceleration means the object slows down until it stops.

Finding final speed

Suppose a trolley starts from rest and accelerates uniformly at $1.5 \mathrm{m/s^2}$ over a distance of $12 \mathrm{m}$. Find its final speed.

Write the known values:

• $u = 0 \mathrm{m/s}$
• $a = 1.5 \mathrm{m/s^2}$
• $s = 12 \mathrm{m}$

Substitute into the equation:

Now take the square root:

The final speed is $6.0 \mathrm{m/s}$. The answer is positive because speed is a size; if the context needs direction, the question will usually make that clear separately.

The common mistake is to stop at $v^2 = 36$ and write $36 \mathrm{m/s}$. That is not a speed. It is the squared speed, so the square root step is essential.

Rearranging the equation

The same relationship can be rearranged to find acceleration or distance. Start from:

Subtract $u^2$ from both sides to get $v^2 - u^2 = 2as$. Then divide both sides by whatever multiplies the unknown.

To find acceleration, divide by $2s$:

To find distance moved, divide by $2a$:

To find initial speed, subtract $2as$ from both sides instead, then take the square root:

For example, a cyclist slows uniformly from $12 \mathrm{m/s}$ to $4.0 \mathrm{m/s}$ over $35 \mathrm{m}$. Determine the acceleration.

The negative sign shows that the acceleration is opposite to the cyclist's motion: the cyclist is slowing down. If a question asks for the deceleration, give the size as $1.8 \mathrm{m/s^2}$, because deceleration already means a slowing-down acceleration.

Signs and stopping distance

Pearson-style calculation marks usually reward the method as well as the final answer. A strong answer shows the equation, substitutes the numbers with units in mind, then gives the numerical answer with the correct unit.

Use this routine:

1. Write $v^2 = u^2 + 2as$.
2. List the known values and the unknown.
3. If the unknown is $a$, $s$ or $u$, rearrange before substituting, or substitute first and then solve, as in the braking example below.
4. Keep the squared speeds as squared speeds until the final square-root step.
5. Check whether acceleration should be positive or negative.
6. Give the final unit: m/s for speed, $\mathrm{m/s^2}$ for acceleration, or m for distance. Round to a sensible number of significant figures, usually the same as the data (often 2).

For a braking calculation, signs matter. A car travelling at $20 \mathrm{m/s}$ brakes uniformly to rest with acceleration $-5.0 \mathrm{m/s^2}$. Its braking distance (the distance moved while the brakes act) is:

This answer is sensible: the distance is positive, while the acceleration is negative because the car is slowing down.

Quick recap

• $v^2=u^2+2as$ applies to constant acceleration and connects motion quantities without using time.
• Find $v^2$ first, then take the square root to obtain speed; squared speed is not the final speed.
• Rearranging gives $a=(v^2-u^2)/(2s)$ or $s=(v^2-u^2)/(2a)$.
• With forward motion positive, braking acceleration is negative. In the general signed equation, $s$ is displacement; it equals forward distance here because there is no reversal.