Kinematics

We start from the definition a = dv/dt and reach a form of acceleration with no time in it at all — useful whenever a question gives you velocity as a function of position.

Step 1 — the definition. Acceleration is the rate of change of velocity with time:

Step 2 — bring in x. The velocity v changes with time only because the particle has moved to a new position x, and x itself changes with time. So v depends on t through x. The chain rule of differentiation lets us insert dx on the top and bottom:

(Think of it as \dfrac{dv}{dt} = \dfrac{dv}{dx}\times\dfrac{dx}{dt} — the two dx factors formally cancel.)

Step 3 — recognise the second factor. By definition dx/dt is the velocity:

Step 4 — substitute. Replacing dx/dt by v:

Result.

What it means. Both forms describe the same acceleration. The first is "how fast does the speed change each second", the second is "how fast does the speed change each metre, multiplied by how many metres you cover per second".

Condition used. The particle must move along a single line (one dimension) and v must be a well-defined function of x. If the particle turns back and passes the same x at two different speeds, you must treat each part of the journey separately.

We start from just two facts about a velocity–time graph — its slope is the acceleration and the area under it is the displacement — and we reach all three equations of uniformly accelerated motion.

Read the graph above: at t = 0 the velocity is u (the intercept), at time t it is v, and because the acceleration is constant the line is straight.

First equation: v = u + at

Step 1. The acceleration is the slope of the straight line. Slope means "rise divided by run". The rise is the velocity gained, v - u; the run is the time taken, t - 0 = t:

Step 2. Multiply both sides by t to clear the fraction:

Step 3. Add u to both sides to make v the subject:

Second equation: s = ut + \frac{1}{2}at^{2}

Step 1. The displacement is the area between the line and the time axis. That shape is a trapezium. Cut it with a horizontal line at height u: below the cut is a rectangle, above it a triangle.

Step 2. The rectangle has height u and width t, so its area is length \times breadth:

Step 3. The triangle has base t and height (v - u), and the area of a triangle is \frac{1}{2}\times base \times height:

Step 4. From the first equation, v - u = at. Substitute it:

Step 5. Add the two areas:

Third equation: v^{2} = u^{2} + 2as

Step 1. Take the trapezium area in one go. The area of a trapezium is \frac{1}{2}\times(sum of the two parallel sides)\times(distance between them). The parallel sides are u and v, and the distance between them is t:

Step 2. We want no t in the answer, so get t from the first equation. Starting from v = u + at, subtract u and divide by a:

Step 3. Substitute this t into the expression for s:

Step 4. Multiply the two brackets. (u+v)(v-u) = v^{2} - u^{2}, the difference-of-squares identity:

Step 5. Multiply both sides by 2a:

Step 6. Add u^{2} to both sides:

Result.

What it means. The first links velocity to time, the second links position to time, and the third links velocity to position with no time involved — so you pick the one that avoids the quantity you were not given.

Condition used. All three need the acceleration a to be constant; that is what makes the graph a straight line. They also need motion in a straight line, with one fixed positive direction chosen before you start. They do not apply to circular motion, or to a car whose acceleration is changing.

We start from the two separate equations of motion, one for the horizontal direction and one for the vertical, and we reach a single equation linking y to x with no time in it. Its shape is what we want to identify.

Step 1 — split the initial velocity. A speed u at angle \theta to the horizontal has a horizontal part and a vertical part, found by resolving:

Step 2 — horizontal motion. There is no horizontal force (we ignore air resistance), so there is no horizontal acceleration and the horizontal velocity stays u\cos\theta for ever. Distance = speed \times time:

Step 3 — vertical motion. Vertically the acceleration is g downwards. Using s = ut + \frac{1}{2}at^{2} with u \to u\sin\theta and a \to -g (taking up as positive):

Step 4 — make t the subject of the horizontal equation. Divide both sides of Step 2 by u\cos\theta:

Step 5 — substitute this t into the vertical equation. This is legal because it is the same t in both — the two motions share one clock:

Step 6 — tidy the first term. The u cancels, and \sin\theta/\cos\theta = \tan\theta:

Step 7 — square the bracket in the second term. Squaring top and bottom:

Step 8 — read off the shape. For a given launch, u and \theta are fixed numbers, so we may write A = \tan\theta and B = g/(2u^{2}\cos^{2}\theta), both constants:

Result. The trajectory equation is

and since y is a quadratic in x with a negative x^{2} coefficient, the path is a parabola opening downwards.

What it means. Gravity pulls the projectile below the straight line it would have followed, and the amount it falls behind grows as t^{2} — that x^{2} term is exactly the free fall the body would have had.

Assumptions used. Air resistance is neglected, g is constant in size and direction over the flight, and the Earth's curvature and rotation are ignored.

We start from the same two motion equations, x = (u\cos\theta)t and y = (u\sin\theta)t - \frac{1}{2}gt^{2}, and reach the three numbers a question always asks for: how long the flight lasts, how high it goes, and how far it lands.

Time of flight T

Step 1. The projectile is back on the ground when its height is zero again, so put y = 0 and t = T:

Step 2. Take out the common factor T:

Step 3. A product is zero when one factor is zero. So either T = 0 (that is the launch instant, which we do not want) or the bracket is zero:

Step 4. Move the second term across:

Step 5. Multiply both sides by 2 and divide by g:

Maximum height H

Step 1. At the highest point the projectile has stopped rising and has not yet started falling, so its vertical velocity is zero there. Apply v^{2} = u^{2} + 2as to the vertical direction only, with initial vertical velocity u\sin\theta, final vertical velocity 0, acceleration -g and displacement H:

Step 2. Write out the squares and the product:

Step 3. Move 2gH to the left:

Step 4. Divide both sides by 2g:

Horizontal range R

Step 1. Horizontally there is no acceleration, so the range is just the constant horizontal speed multiplied by the whole time of flight:

Step 2. Put in T = 2u\sin\theta/g from above:

Step 3. Multiply the two u factors together:

Step 4. Use the double-angle identity 2\sin\theta\cos\theta = \sin 2\theta:

Maximum range

Step 1. In R = u^{2}\sin 2\theta/g, only \sin 2\theta can change if u is fixed. The largest value of a sine is 1:

Step 2. This happens when \sin 2\theta = 1, i.e. when 2\theta = 90^{\circ}:

Result.

What it means. Time of flight and height depend only on the vertical component u\sin\theta; the range needs both components, which is why there is a best compromise angle of 45^{\circ}. Note also T = 2\times(time to reach the top), so rise time equals fall time.

Assumptions used. No air resistance, constant g, and — important — the landing point is at the same height as the launch point. If the ground is lower (a ball thrown off a cliff), T and R must be recalculated from y = -h instead of y = 0.