Calculating Acceleration and Uniform Motion

What acceleration tells us

Velocity describes both how fast an object is moving and its direction. Acceleration describes how quickly that velocity changes. An object accelerates if it speeds up, slows down or changes direction, although this lesson concentrates on motion along a straight line.

Acceleration is measured in metres per second squared, $\text{m/s}^2$. An acceleration of $3\text{ m/s}^2$ means that the velocity changes by $3\text{ m/s}$ during each second, provided the acceleration remains uniform.

Near the Earth's surface, a freely falling object has an acceleration of about $9.8\text{ m/s}^2$. This gives a useful scale for estimates: an acceleration of a few $\text{m/s}^2$ is plausible for many ordinary vehicles, whereas some rapidly launched objects can have accelerations of tens of $\text{m/s}^2$.

Calculating average acceleration

Average acceleration is the change in velocity divided by the time taken:

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

Here, $a$ is acceleration in $\text{m/s}^2$, $u$ is initial velocity in , is in and is time in seconds. The symbol means ‘change in’, so : final velocity minus initial velocity.

For example, a race car that changes velocity from $12\text{ m/s}$ to $48\text{ m/s}$ in $6.0\text{ s}$ has

$$a=\frac{48-12}{6.0}=6.0\text{ m/s}^2.$$

This is an average because it describes the velocity change across the whole six-second interval. It does not by itself prove that the acceleration had the same value at every instant.

Direction, signs and deceleration

Choose one direction as positive before using velocity in a calculation. Motion in the opposite direction then has a negative velocity.

If an object moving in the positive direction slows down, $v-u$ is negative, so its acceleration is negative. For example, changing from $20\text{ m/s}$ to $5\text{ m/s}$ gives a negative change in velocity. The object is decelerating because its speed is decreasing.

Negative acceleration does not always mean slowing down. An object travelling in the negative direction can speed up while having negative acceleration. The reliable rule is:

• velocity and acceleration in the same direction: speed increases;
• velocity and acceleration in opposite directions: speed decreases.

In everyday language, a ‘deceleration of $5\text{ m/s}^2$’ often means a deceleration magnitude of $5\text{ m/s}^2$. In a signed calculation, it would be written as $a=-5$ if the original motion is positive.

Uniform acceleration without time

When acceleration is uniform, the following equation connects initial velocity, final velocity, acceleration and distance:

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

The symbols and units are:

• $v$: final velocity, in $\text{m/s}$;
• $u$: initial velocity, in $\text{m/s}$;
• $a$: uniform acceleration, in ;

This equation is especially useful when time is not given. It must not be applied to a whole journey whose acceleration varies unless the question provides a suitable constant acceleration for the interval being considered.

To find final velocity, first rearrange to

$$v^2=u^2+2as.$$

Squaring removes information about direction, so taking the square root gives two mathematical possibilities:

$$v=\pm\sqrt{u^2+2as}.$$

Choose the sign using the stated positive direction and the physical context. If the object is still moving in the chosen positive direction, use the positive root; if it is moving in the opposite direction, use the negative root. If the question asks only for final speed, give the non-negative magnitude $\sqrt{u^2+2as}$.

For example, if a motorcycle is known to continue forwards after accelerating, the forward direction is positive and the positive root is appropriate. The equation alone, however, cannot distinguish a velocity of $+20\text{ m/s}$ from one of $-20\text{ m/s}$ because both square to the same value.

If an object starts from rest, $u=0$, so the relationship simplifies to $v^2=2as$; the direction or context is still needed when stating a signed velocity.

Rearranging the uniform-acceleration equation

The same equation can answer different questions. To find acceleration,

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

To find distance,

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

Suppose a motorcycle travelling at $25\text{ m/s}$ comes to rest with a uniform acceleration of $-5.0\text{ m/s}^2$. Taking its original direction as positive gives $u=25\text{ m/s}$, and . Therefore,

$$s=\frac{0^2-25^2}{2(-5.0)}=62.5\text{ m}.$$

The two negative signs cancel, producing a positive distance. This also shows why signs should be retained rather than replacing acceleration with an unsigned number halfway through a calculation.

Choosing the appropriate equation

First list the quantities given and the quantity required. Use

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

when the initial velocity, final velocity and time are involved. Use

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

when acceleration is uniform and the relevant quantities are velocity, acceleration and distance, with no need for time.

Finally, attach the correct unit and interpret the result. A value of $2.5\text{ m/s}^2$ means the velocity changes by an average of $2.5\text{ m/s}$ each second; it does not mean that the object is travelling at $2.5\text{ m/s}$.

Average acceleration

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