Motion

Before we can describe motion precisely, we must fix a reference point (or origin) against which the position of an object is measured. An object is said to be in motion if its position changes with time relative to this reference point, and at rest if it does not.

To describe how far an object has moved, we use two related but different quantities:

• Distance is the total length of the actual path covered by an object, irrespective of direction. It is a scalar quantity (only magnitude, no direction) and is always positive.
• Displacement is the shortest straight-line distance between the initial position and the final position of the object, together with the direction from the initial to the final point. It is a vector quantity, and its magnitude is always less than or equal to the distance travelled.

Worked example: Aria walks 4 m due East and then 3 m due North. Find the distance and the displacement covered by her.

• Distance = 4 m + 3 m = 7 m (sum of the actual path lengths walked).
• Displacement = shortest straight line from the start point to the end point. Since the two legs are perpendicular: displacement = √(4² + 3²) = √(16 + 9) = √25 = 5 m, directed from the starting point to the final point.

Notice that displacement (5 m) is smaller than distance (7 m). If Aria had instead walked in a closed loop and returned to her starting point, her distance travelled would still be positive, but her displacement would be zero.

Motion can also be classified by how the distance changes with time:

• Uniform motion: the object covers equal distances in equal intervals of time, however small the intervals may be (e.g. a car moving at a constant 40 km/h on a straight highway).
• Non-uniform motion: the object covers unequal distances in equal intervals of time (e.g. a bus slowing down and speeding up in city traffic).

Speed tells us how fast an object is moving, while velocity tells us how fast it is moving and in which direction.

• Speed is the distance travelled by an object per unit time. It is a scalar quantity: v = s/t.
• Velocity is the displacement of an object per unit time. It is a vector quantity, since displacement itself is a vector.

When an object's speed (or velocity) keeps changing, we describe its overall motion using average speed and average velocity.

• Average speed = total distance travelled ÷ total time taken.
• Average velocity = total displacement ÷ total time taken. If velocity changes at a uniform rate from an initial value u to a final value v, average velocity = (u + v)/2.

Worked example: A car travels at 40 km/h for the first 15 minutes and then at 60 km/h for the next 15 minutes. Find its average speed for the whole journey.

• Time for each stage = 15 min = 0.25 h.
• Distance in stage 1 = speed × time = 40 km/h × 0.25 h = 10 km.
• Distance in stage 2 = 60 km/h × 0.25 h = 15 km.
• Total distance = 10 km + 15 km = 25 km. Total time = 0.25 h + 0.25 h = 0.5 h.
• Average speed = total distance ÷ total time = 25 km ÷ 0.5 h = 50 km/h.

Here the answer equals the simple arithmetic mean of 40 and 60 only because the car spent equal time at each speed; if it had covered equal distances at each speed instead, the average speed would be the harmonic mean, 2v₁v₂/(v₁+v₂) = 48 km/h, which is different.

An object's velocity rarely stays constant for long — it speeds up, slows down, or changes direction. Acceleration is the rate of change of velocity with time.

If an object's velocity changes from an initial value u to a final value v in time t, its acceleration is a = (v − u)/t.

• If velocity increases with time, acceleration is positive.
• If velocity decreases with time, acceleration is negative — this is also called retardation or deceleration.
• If velocity changes by equal amounts in equal intervals of time, the motion has uniform acceleration (e.g. a freely falling object). If it changes by unequal amounts, the acceleration is non-uniform.

Worked example: A car's velocity increases uniformly from 5 m/s to 25 m/s in 10 s. Find its acceleration.

• Given: u = 5 m/s, v = 25 m/s, t = 10 s.
• Formula: a = (v − u)/t.
• Substitution: a = (25 − 5)/10 = 20/10.
• Result: a = 2 m/s², meaning the velocity increases by 2 m/s every second.

If instead a train slows from 25 m/s to 5 m/s in 10 s, a = (5 − 25)/10 = −2 m/s². The negative sign shows a retardation of magnitude 2 m/s².

Graphs let us "see" motion at a glance and read off speed, velocity, acceleration, and distance without long calculations.

Distance-time graphs plot distance covered (y-axis) against time (x-axis).

• A straight line means the object covers equal distances in equal times, i.e. it moves with uniform speed. The slope of this line equals the speed.
• A curved line means the speed is changing, i.e. the motion is non-uniform.
• A horizontal line (zero slope) means the object is at rest.

Velocity-time graphs plot velocity (y-axis) against time (x-axis).

• The slope of a velocity-time graph gives the acceleration of the object.
• The area enclosed between the velocity-time graph and the time axis gives the distance (or displacement) covered in that time interval.
• A line parallel to the time axis represents motion with uniform velocity (zero acceleration).

Worked example: An object starts from rest and accelerates uniformly at 2 m/s² for 5 s. Use the velocity-time graph to find the distance it covers.

• At t = 0, velocity u = 0. At t = 5 s, velocity v = u + at = 0 + 2 × 5 = 10 m/s.
• The velocity-time graph is a straight line from (0, 0) to (5 s, 10 m/s), so the region under it is a right-angled triangle.
• Area of triangle = ½ × base × height = ½ × (5 s) × (10 m/s) = 25 m.
• So the object covers 25 m in the 5 seconds — the same value obtained from s = ut + ½at².

For an object moving with uniform acceleration a, starting with initial velocity u and reaching final velocity v after time t, the velocity-time graph is a straight line. Three useful equations can be read directly from this graph.

• First equation (from the slope): since acceleration = slope = (v − u)/t, rearranging gives v = u + at.
• Second equation (from the area under the graph): the area under the line (a trapezium, split into a rectangle of height u and a triangle above it) gives the distance s travelled: s = ut + ½at².
• Third equation (eliminating time): combining the two equations above to remove t gives v² = u² + 2as, useful when time is not known or not needed.

Worked example: A scooter starts from rest and accelerates uniformly at 5 m/s² for 4 s. Find its final velocity and the distance covered, and check the results using the third equation.

• Given: u = 0, a = 5 m/s², t = 4 s.
• v = u + at = 0 + 5 × 4 = 20 m/s.
• s = ut + ½at² = 0 + ½ × 5 × 4² = 0.5 × 5 × 16 = 40 m.
• Check: v² = u² + 2as → 20² = 0² + 2 × 5 × 40 → 400 = 400. The equations agree.

Now consider an object moving at constant speed along a circular path — this is called uniform circular motion. Even though the speed does not change, the direction of motion changes continuously at every point on the circle, so the velocity (a vector) is continuously changing. Since velocity is changing, the object is accelerating, even though its speed stays constant.

Worked example: An object moves once around a circular track of radius 7 m in 22 s. Find its speed.

• Given: radius r = 7 m, time for one round T = 22 s, take π ≈ 22/7.
• Circumference (distance for one full round) = 2πr = 2 × (22/7) × 7 = 44 m.
• Speed = distance/time = 44 m ÷ 22 s = 2 m/s.

Although the object's speed stays at 2 m/s throughout, its velocity keeps changing direction, so uniform circular motion is an example of accelerated motion at constant speed.