Kinematics

Distance, displacement, speed, velocity and acceleration

Every quantity in kinematics comes from asking one of two questions: how far, or how fast is that distance changing. Distance is the total length of the path you actually travel and it has no direction: walk 3 m to the door and 3 m back and your distance is 6 m. Displacement is how far you end up from where you started, measured in a straight line and with the direction stated: after that same walk your displacement is 0 m, because you are back where you began. Distance is a scalar; displacement is a vector.

The same split runs through the rates. Speed is the rate of change of distance and is a scalar - a car's speedometer reads 60 mph whatever direction the car is pointing. Velocity is the rate of change of displacement, v = delta s / delta t, and it is a vector, so a change of direction is a change of velocity even at constant speed. A car going round a roundabout at a steady 20 m s-1 has constant speed but a changing velocity, and it is therefore accelerating.

Acceleration is the rate of change of velocity, a = delta v / delta t, measured in m s-2. Notice that the definition uses velocity, not speed: an object accelerates if its speed changes, if its direction changes, or both. A negative value of a does not mean the object is slowing down - it means the acceleration points the opposite way to the positive direction you chose, so an object can have a negative acceleration and still be speeding up.

• Distance: a scalar, the length of path travelled, measured in metres, never negative and never decreasing.
• Displacement: a vector, the straight-line distance from start to finish with its direction, measured in metres.
• Speed: a scalar, the rate of change of distance, measured in m s-1.
• Velocity: a vector, the rate of change of displacement, measured in m s-1.
• Acceleration: a vector, the rate of change of velocity, measured in m s-2.

Displacement-time graphs: the gradient is the velocity

A graph of displacement against time tells you everything about the motion if you read it the right way round. The gradient of a displacement-time graph is the velocity: a steep line means a large velocity, a shallow line a small one, and the sign of the gradient gives the direction of travel. A horizontal line means the object is stationary. The height of the graph only tells you where the object is, never how fast it is going.

Where the graph is curved the velocity is changing, and the velocity at a particular instant is the gradient of the tangent drawn at that point. A curve that is getting steeper means the object is speeding up, and a curve that is flattening means it is slowing down. The average velocity over a section is the gradient of the straight line joining the two ends of that section, not the gradient of the tangent.

The speed is the magnitude of the gradient, so it is always positive, while the velocity may be positive or negative. If the graph crosses the time axis the object has reversed its direction, and at a turning point of the curve, where the gradient is zero, the object is momentarily at rest.

• Straight horizontal line: the object is stationary, so the velocity is zero.
• Straight sloping line: constant velocity, equal to the gradient of the line.
• Curved line: the velocity is changing, and the gradient of the tangent gives the instantaneous velocity.

Velocity-time graphs: gradient is acceleration, area is displacement

A velocity-time graph gives two pieces of information at once and you must be able to use both. The gradient is the acceleration, and the area between the graph and the time axis is the displacement. The units confirm it: a gradient of m s-1 divided by s gives m s-2, and an area of m s-1 multiplied by s gives m.

The shapes are worth learning. A horizontal line is constant velocity and zero acceleration. A straight sloping line is uniform, that is constant, acceleration. A curve means the acceleration itself is changing, and the instantaneous acceleration is the gradient of the tangent at that point. A line crossing the time axis shows that the object has reversed its direction.

To find areas, split the graph into triangles and rectangles, or count squares on graph paper when the line is curved. Areas below the time axis count as negative displacement, because the object has moved back past its starting point, but the total distance travelled is the sum of the sizes of all the areas with every one treated as positive. Worked case: a cyclist accelerates from rest to 12 m s-1 in 4.0 s, rides at a steady 12 m s-1 for 10 s, then brakes to rest in 6.0 s. The acceleration is 12 / 4.0 = 3.0 m s-2, the deceleration is 12 / 6.0 = 2.0 m s-2, and the displacement is (0.5 x 4.0 x 12) + (10 x 12) + (0.5 x 6.0 x 12) = 24 + 120 + 36 = 180 m.

Deriving the equations of motion

The equations for uniformly accelerated motion in a straight line are not magic; they come straight out of the two definitions you already have. Start with acceleration, a = (v - u) / t, where u is the initial velocity and v the final velocity. Rearranging gives v = u + at, the first equation.

Next, displacement. If the acceleration is uniform then the velocity rises steadily, so the average velocity is (u + v) / 2 and the displacement is s = ((u + v) / 2) x t, that is s = (u + v)t / 2. This is exactly the area under the velocity-time graph, read as the area of a trapezium.

Now eliminate one quantity at a time. Substituting v = u + at into s = (u + v)t / 2 gives s = (2u + at)t / 2, which is s = ut + (1/2)at^2. Rearranging the first equation to t = (v - u) / a and substituting that into s = (u + v)t / 2 gives s = (u + v)(v - u) / 2a, and multiplying by 2a gives 2as = v^2 - u^2, so v^2 = u^2 + 2as. A fifth form, s = vt - (1/2)at^2, comes from putting u = v - at into the third equation, and it is the one to use when you know the final velocity but not the initial one.

Every one of these equations assumes that the acceleration is constant in both magnitude and direction, that the motion is in a straight line, and that all five symbols refer to the same interval of time. For a curved path such as a projectile you cannot use them directly: you must split the motion into two perpendicular components and apply the equations to each component separately.

Solving problems with the equations of motion

Most marks in these questions are lost for signs and units rather than for physics. Choose a positive direction at the start of the question and write it down. For a ball thrown upwards, taking upwards as positive makes a = -9.81 m s-2, while taking downwards as positive makes a = +9.81 m s-2. Either choice works, but you must keep it for the whole question, and any velocity in the opposite direction then carries a minus sign.

Write out the five quantities s, u, v, a and t, fill in what you know, and mark the one you want. That identifies the equation to use immediately: if the time is missing, use v^2 = u^2 + 2as, and if the displacement is missing, use v = u + at. Never mix values from different stages of a journey in one calculation. Split the journey into stages, apply the equations to each stage separately, and carry the final velocity of one stage forward as the initial velocity of the next.

Worked example. A ball is thrown vertically upwards from ground level at 15 m s-1, and air resistance is negligible. Take upwards as positive, so u = +15 m s-1, v = 0 at the highest point and a = -9.81 m s-2. The time to the highest point is t = (v - u) / a = (0 - 15) / (-9.81) = 1.53 s, and the greatest height is s = (v^2 - u^2) / 2a = (0 - 225) / (2 x -9.81) = 11.5 m. The ball then falls back through the same 11.5 m, and the speed with which it returns to the ground equals the speed at which it was thrown, 15 m s-1, so the total time of flight is 2 x 1.53 = 3.1 s.

Two reminders. Acceleration is a vector, so a deceleration of 2.0 m s-2 must be entered as a = -2.0 m s-2 when the direction of travel is positive. And keep the significant figures sensible: if the data are given to two significant figures, quote your answer to two, and always attach the unit.

• State your positive direction before you substitute anything, and keep it throughout.
• List s, u, v, a and t; the missing quantity tells you which equation to use.
• Treat each stage of a journey separately, but note that v for one stage is u for the next.
• Give every answer with its unit, and match the number of significant figures to the data.

Free fall and measuring the acceleration of free fall

Free fall is motion under the force of gravity alone, with no other force acting. In a uniform gravitational field every object falls with the same acceleration, g = 9.81 m s-2 downwards, whatever its mass: a heavy ball and a light ball released together from the same height reach the ground at the same time, because the larger weight on the heavier ball acts on a proportionally larger mass. Air resistance spoils the experiment in the laboratory, since the drag depends on shape and speed rather than on mass, and a light, fluffy object falls visibly more slowly. Remove the air and the equality is exact: on the Apollo 15 mission in 1971 Commander David Scott dropped a geological hammer and a falcon's feather together on the Moon and they landed at the same moment.

To measure g, start from the free-fall equation with u = 0. Since s = ut + (1/2)at^2 becomes h = (1/2)gt^2, we get g = 2h / t^2. Hold a steel ball bearing with an electromagnet, and switch the current off to release it; the same switch starts an electronic timer. The ball falls a measured height h and breaks a light beam, or strikes a trapdoor, to stop the timer. The electromagnet matters because it removes the reaction-time delay that would be there if the ball were released by hand.

One reading is not enough. A time of about a second measured with a stopwatch carries a large percentage uncertainty, so repeat the drop several times at the same height and average, then repeat the whole experiment for a range of heights. Plot a graph of h against t^2: because h = (1/2)gt^2, this is a straight line through the origin of gradient g/2, so g = 2 x gradient. The graph spreads the error over many readings and turns a random timing error into a scatter you can actually see.

Other methods rest on the same physics. Two light gates a known distance apart time the ball between them, and a single light gate with a card of measured length gives a velocity each time it passes. A stroboscope flashing at a fixed rate, or a video camera recording at a known frame rate, lets you measure the position of a falling object frame by frame, and a graph of displacement against t^2 again has gradient g/2. Whatever the method, use a small, dense object so that air resistance is negligible, measure h from the bottom of the ball to the trapdoor, and look straight at the scale to avoid a parallax error.

• Apparatus: electromagnet, steel ball bearing, metre rule, electronic timer with a trapdoor or light gate.
• Measure the height h and the time t of the fall, then calculate g = 2h / t^2.
• Repeat at each height and average, then plot h against t^2 and use g = 2 x gradient.
• Reduce error with a small dense ball and an electromagnet; avoid parallax when reading the scale.

Projectile motion: uniform velocity one way, uniform acceleration the other

A projectile is an object moving under gravity alone after it has been given an initial velocity. Its motion splits into two independent parts. Horizontally there is no force at all if air resistance is negligible, so the horizontal velocity is constant and the horizontal displacement is u_x multiplied by t. Vertically the only force is the weight, so the object accelerates downwards at g = 9.81 m s-2 and the vertical motion is simply free fall.

The two parts do not affect each other, and that is the idea examiners test. Project a ball horizontally from a table and drop an identical ball from the same height at the same instant, and the two reach the floor together: the horizontal velocity has no vertical component, so it cannot change how long the fall takes. Projecting the ball changes where it lands and not when.

The standard method is to find the time from the vertical motion and then use that time in the horizontal motion. Worked example: a ball is thrown horizontally at 12 m s-1 from the top of a cliff 45 m above the sea. Vertically u = 0, s = 45 m and a = 9.81 m s-2, so 45 = (1/2) x 9.81 x t^2, giving t = square root of (2 x 45 / 9.81) = 3.03 s. Horizontally the velocity stays at 12 m s-1, so the horizontal distance is 12 x 3.03 = 36 m. At the moment of landing the velocity has two components, a horizontal 12 m s-1 and a vertical v = gt = 9.81 x 3.03 = 29.7 m s-1. Combining them with Pythagoras gives a speed of 32 m s-1 at about 68 degrees below the horizontal.

A launch at an angle needs exactly the same treatment. Resolve the initial velocity into u cos theta horizontally and u sin theta vertically, then follow each direction separately: the vertical velocity decreases on the way up and increases again on the way down, while the horizontal velocity is unchanged at u cos theta. Because g is constant, the vertical half of the motion is uniformly accelerated and the equations of motion apply to it directly, with a = 9.81 m s-2 downwards.