Displacement

Displacement describes an object’s change in position, including its direction. It depends only on where the object starts and where it finishes, not on the path it takes between those positions.

Displacement is a vector: it has both magnitude and direction. Its unit is the metre ($$\mathrm{m}$$).

Displacement along a straight line

Choose a positive direction and measure the initial and final positions from the same reference point. The displacement is the final position minus the initial position:

Here, $$\Delta s$$ is displacement, while $$s_{\text{initial}}$$ and $$s_{\text{final}}$$ are the initial and final positions. A positive displacement is in the chosen positive direction; a negative displacement is in the opposite direction.

Displacement and distance travelled

Distance travelled is the total length of the path taken. It is a scalar: it has magnitude but no direction. If an object changes direction, its displacement can have a smaller magnitude than the distance it travels.

For example, suppose an object moves $$5,\mathrm{m}$$ east and then $$2,\mathrm{m}$$ west. Its distance travelled is $$7,\mathrm{m}$$. It finishes $$3,\mathrm{m}$$ east of its starting point, so its displacement is $$3,\mathrm{m}$$ east, or $$+3,\mathrm{m}$$ if east is chosen as positive.

If the object returns to its starting point, its displacement is zero, even though it has travelled a distance.

Speed and Velocity

Speed describes how quickly an object travels. It is a scalar: it has a magnitude, but no direction. Velocity describes how quickly an object’s position changes and in which direction. It is a vector: it has both magnitude and direction.

Both speed and velocity are measured in metres per second ($$\mathrm{m/s}$$).

Average speed and average velocity

Distance is the total length of the path travelled. Average speed is the total distance travelled divided by the time taken:

Displacement is the change in position from the starting point to the finishing point, including its direction. Average velocity is the displacement divided by the time taken:

Here, $$v$$ is average velocity, $$\Delta s$$ is displacement and $$\Delta t$$ is the time interval. A velocity can be positive or negative depending on the chosen positive direction. For example, if east is positive, a velocity towards the west is negative.

Average speed and the magnitude of average velocity need not be equal when an object changes direction. For example, suppose an object travels $$10\ \mathrm{m}$$ east and then $$10\ \mathrm{m}$$ west, taking $$10\ \mathrm{s}$$ in total. Its distance travelled is $$20\ \mathrm{m}$$, so its average speed is $$2\ \mathrm{m/s}$$. Its displacement is zero because it finishes where it started, so its average velocity is $$0\ \mathrm{m/s}$$.

Instantaneous speed and velocity

Average values describe motion over an interval. Instantaneous speed and instantaneous velocity describe motion at a particular instant.

Instantaneous velocity gives the speed and direction at that instant. Instantaneous speed is the magnitude of instantaneous velocity, so it is not negative. For example, if the instantaneous velocity is $$-3\ \mathrm{m/s}$$, the instantaneous speed is $$3\ \mathrm{m/s}$$.

Acceleration

An object accelerates when its velocity changes. The change may be in its speed, its direction of motion, or both. Acceleration describes how quickly velocity changes.

Average acceleration

Over a time interval, average acceleration is the change in velocity divided by the time taken:

Here, $$a$$ is average acceleration, $$\Delta v$$ is the final velocity minus the initial velocity, and $$\Delta t$$ is the elapsed time. Acceleration is measured in metres per second squared, written $$\mathrm{m/s^2}$$.

Acceleration is a vector, so it has a direction as well as a magnitude. For straight-line motion, choose one direction as positive; acceleration in the opposite direction is negative. A negative acceleration does not always mean that an object is slowing down.

For example, suppose right is positive and an object’s velocity changes from $$+6\ \mathrm{m/s}$$ to $$+2\ \mathrm{m/s}$$ in $$2\ \mathrm{s}$$. Its average acceleration is $$(-4\ \mathrm{m/s})/(2\ \mathrm{s}) = -2\ \mathrm{m/s^2}$$. The acceleration is to the left, opposite to the object’s velocity, so the object slows down.

Acceleration and direction of motion

In straight-line motion, the effect on speed depends on how the direction of acceleration compares with the direction of velocity:

• Acceleration in the same direction as velocity increases speed.
• Acceleration in the opposite direction to velocity decreases speed.

If acceleration continues opposite to the motion until the object stops, it can then make the object move in the opposite direction. Acceleration can therefore change the direction of motion as well as the speed.

Uniform and non-uniform acceleration

Uniform acceleration is constant: velocity changes by equal amounts in equal time intervals. Non-uniform acceleration changes with time, so the change in velocity need not be equal in equal time intervals.

Graphs of Straight-Line Motion

Motion along a straight line can be described using displacement, velocity and acceleration. A graph shows how one of these quantities changes with time. Its gradient or signed area can then reveal another quantity.

Because displacement, velocity and acceleration have direction, one direction is chosen as positive. Values in the opposite direction are negative.

Gradients: velocity and acceleration

The gradient of a displacement–time graph gives velocity. Over an interval, the gradient of the straight line joining its endpoints gives the average velocity:

A positive gradient means motion in the positive direction; a negative gradient means motion in the negative direction. A horizontal section has zero gradient, so the object is stationary during that section.

The gradient of a velocity–time graph gives acceleration. Over an interval, the gradient of the straight line joining its endpoints gives the average acceleration:

For either graph, the gradient of a tangent at a particular point gives the instantaneous rate: velocity on a displacement–time graph or acceleration on a velocity–time graph.

A straight velocity–time graph has a constant gradient, so acceleration is uniform. A curved velocity–time graph has a changing gradient, so acceleration is non-uniform.

Signed areas: displacement and change in velocity

The signed area between a velocity–time graph and the time axis gives the displacement during that time interval. Area above the time axis contributes positively; area below it contributes negatively. Add these contributions with their signs to find the net displacement.

Displacement is not necessarily the total distance travelled. If the object reverses direction, a negative contribution can cancel some of a positive contribution. Total distance instead includes the magnitudes of the areas on both sides of the axis, without cancellation.

Similarly, the signed area between an acceleration–time graph and the time axis gives the change in velocity. An area below the axis represents a negative change in velocity.

Motion graphs for a bouncing ball

Take upward as the positive direction. Just before the ball hits the ground, its velocity is negative because it is moving downward. After the bounce, its velocity is positive because it is moving upward: the sign change shows that the ball has reversed direction.

Between bounces, gravity accelerates the ball downward. On a velocity–time graph, its velocity therefore decreases as time passes: it becomes more negative while the ball falls, and becomes less positive while the ball rises.

During contact with the ground, the ball’s velocity changes rapidly from negative to positive. On the velocity–time graph, this corresponds to a steep positive gradient and hence a large upward acceleration over a short time. The signed area under the acceleration–time graph during contact gives this change in velocity.

Equations for Uniform Acceleration

When an object moves in a straight line with constant acceleration, its velocity changes by the same amount in each equal interval of time. This makes it possible to relate its displacement, elapsed time and velocities using the equations below.

The symbols refer to the same interval of motion:

• $$u$$: initial velocity
• $$v$$: final velocity
• $$a$$: constant acceleration
• $$t$$: elapsed time
• $$s$$: displacement

The first equation says that the change in velocity is acceleration multiplied by time. Because the acceleration is constant, the average velocity over the interval is $$\frac{u+v}{2}$$. Multiplying this average velocity by time gives the second equation for displacement.

The third equation combines these relationships to give displacement without needing the final velocity. The fourth relates velocity and displacement without needing the time.

Choosing an acceleration equation

Identify the quantities you know and the quantity you need. Choose an equation that contains those quantities but does not require another unknown. For example, if initial velocity, acceleration and time are known, use $$s = ut + \frac{at^2}{2}$$ to find displacement without first finding final velocity.

Use consistent units, such as metres, seconds, metres per second and metres per second squared. Choose one direction along the line as positive: displacement, both velocities and acceleration must all use that same choice. A negative value indicates a direction opposite to the chosen positive direction.

Uniform acceleration: worked calculation

An object initially moves at $$10\ \mathrm{m/s}$$ in the positive direction. Its acceleration is $$-2\ \mathrm{m/s^2}$$ for $$3\ \mathrm{s}$$. To find its displacement, use the equation that contains $$u$$, $$a$$, $$t$$ and $$s$$:

The displacement is $$21\ \mathrm{m}$$ in the positive direction. The acceleration is negative because it points opposite to that direction.

These equations require constant acceleration throughout the interval; they do not generally apply over an interval in which acceleration varies.

Acceleration Due to Gravity

An object in free fall accelerates because gravity acts on it. Near Earth’s surface, this acceleration has a magnitude of approximately $$9.81\ \text{m s}^{-2}$$ and is directed downwards. Its magnitude is denoted by $$g$$.

When air resistance is negligible, the acceleration is approximately constant. This means the object’s downward velocity changes by about $$9.81\ \text{m s}^{-1}$$ each second. The acceleration remains downwards whether the object is moving up or down.

Upward motion and the highest point

As an object moves upwards, gravity accelerates it downwards, so its upward velocity decreases. At the highest point, its velocity is momentarily zero, but its acceleration is still approximately $$9.81\ \text{m s}^{-2}$$ downwards. The object then moves downwards, gaining downward speed.

Sign of gravitational acceleration

The sign used in a calculation depends on the chosen positive direction. Since gravity accelerates the object downwards:

• If upwards is positive, gravitational acceleration is approximately $$-g = -9.81\ \text{m s}^{-2}$$.
• If downwards is positive, gravitational acceleration is approximately $$+g = +9.81\ \text{m s}^{-2}$$.

The physical acceleration is the same in both cases; only its sign changes with the convention.