2. Kinematics

Distance is total path length, a scalar. Displacement is the directed change from initial to final position, a vector. Speed is distance per time; velocity is displacement per time.

Choose a sign convention for one-dimensional motion and distinguish average quantities from instantaneous derivatives.

A runner completes one 400 m lap and returns to the start: distance is 400 m but displacement is zero.

Zero displacement does not mean the object never moved, and speed cannot be negative.

Velocity v=ds/dt is the rate of change of displacement; acceleration a=dv/dt is the rate of change of velocity. Their signs depend on the chosen direction.

A negative velocity can mean motion opposite to the positive axis, while negative acceleration does not automatically mean slowing down—compare signs of v and a.

If v is positive and a negative, the object slows until v reaches zero; if both are negative, speed can increase.

Acceleration opposite to velocity means slowing only while the velocity remains in that direction.

Displacement over an interval is the integral ∫v dt, represented by the signed area under a velocity–time graph. Areas below the time axis are negative.

Split the graph at zero crossings or changes of shape, then add signed areas. Total distance uses absolute areas instead.

A triangular v–t graph with base 4 s and height 6 m s⁻¹ gives displacement 12 m if it stays above zero.

The area gives displacement, not automatically distance; a return trip can cancel signed areas.

The tangent gradient ds/dt at a point on an s–t graph is velocity. A secant gradient over an interval is average velocity.

Read the sign from the slope and compare steepness for speed. A horizontal tangent means instantaneous velocity zero, not necessarily no future motion.

For s=t², the tangent gradient at t=3 is 6 m s⁻¹, while the average velocity from t=2 to 3 is 5 m s⁻¹.

The height of an s–t graph is position; velocity is its gradient.

Acceleration is the tangent gradient dv/dt on a velocity–time graph. A straight segment has constant acceleration; a horizontal segment has zero acceleration.

Use signed gradients and identify the interval. A negative gradient means velocity decreases in the chosen coordinate direction, not necessarily negative speed.

A velocity falling from 10 to 4 m s⁻¹ over 3 s has average acceleration −2 m s⁻²; the tangent gives instantaneous values if the curve is non-linear.

The area under a v–t graph is displacement, while its gradient is acceleration—do not interchange them.

If acceleration a is constant, integrating a=dv/dt gives v=u+at; integrating v=ds/dt then gives s=ut+½at² and v²=u²+2as.

The derivation assumes constant a and consistent initial conditions. Use the equation containing the known quantities and keep displacement signed.

Eliminating t from v=u+at and s=½(u+v)t yields v²=u²+2as.

These equations cannot be used unchanged when acceleration varies with time or position.

For constant acceleration, v=u+at, s=ut+½at², v²=u²+2as and s=½(u+v)t connect displacement, velocity, time and acceleration.

Define positive direction, use signed variables and choose the equation containing the known quantities. Split a journey when acceleration changes.

A car changing from 5 to 17 m s⁻¹ in 4 s has a=3 m s⁻² and displacement 44 m under the constant-a model.

SUVAT is not a universal motion formula set; variable acceleration needs calculus or graph methods.

Near Earth, a falling object has approximately constant downward acceleration g when air resistance is negligible. Measure distance and time, then use s=½gt² from rest or a suitable model.

Repeat timings, reduce reaction-time error, use a light gate or video where possible, and plot quantities that test the assumed relationship.

A plot of fall distance s against t² has gradient g/2, so g is twice the fitted gradient.

The measured value can differ from 9.81 because of timing, height, drag or calibration; that does not redefine g.

For a projectile, horizontal velocity remains constant while vertical acceleration is −g. Position is described by x=u_xt and y=u_yt−½gt².

Solve components with the same time variable, then eliminate time only after applying launch and landing conditions. Air resistance is excluded by the ideal model.

A horizontal launch from height h has flight time √(2h/g) and horizontal range u√(2h/g).

The horizontal and vertical motions share time but not acceleration; gravity does not create horizontal acceleration.