A.1 Kinematics

Track change in position

Velocity $\vec v$ describes how position changes with time. It is a vector: its direction is the direction of motion at that instant, and on a curved path the velocity arrow is tangent to the path.

Track change in velocity

Acceleration $\vec a$ describes how velocity changes with time. A change in speed, direction, or both is acceleration; an object can accelerate even while its instantaneous speed is zero.

Acceleration changes velocity

Acceleration is the rate of change of velocity:

$$\vec a=\frac{d\vec v}{dt}$$

A constant acceleration gives equal changes in velocity during equal time intervals.

Read the gradient

On a position–time graph, the gradient represents velocity. On a velocity–time graph, the gradient represents acceleration. The graph’s slope, not its height alone, carries the rate-of-change meaning.

Ignore the route for displacement

The path taken between the two positions does not determine displacement. A curved or complicated journey can still have a straight-line displacement between its endpoints.

Use components when needed

For perpendicular changes, resolve displacement into components and combine them vectorially. A signed one-dimensional displacement is positive or negative according to the chosen axis.

Displacement

Displacement is the vector change in position from start to finish. Its magnitude is the shortest endpoint-to-endpoint separation, not generally the length of the route.

Match the average quantity

Average speed uses total distance divided by total time. Average velocity uses displacement divided by total time. A route with turns can therefore have average speed greater than the magnitude of average velocity.

Instantaneous values use one moment

Instantaneous velocity is the tangent gradient on a position–time graph; instantaneous speed is its magnitude and is what an ideal speedometer reports. Instantaneous acceleration is the tangent gradient on a velocity–time graph. Average values instead use a finite interval.

Connect graph quantities

The gradient of a velocity–time graph is acceleration, and the area under it is displacement. A constant acceleration therefore produces a straight-line velocity–time graph.

Choose an equation

List the known and unknown quantities $s,u,v,a,t$. Select an equation containing the required unknown and only known quantities; keep signs consistent with the positive direction.

Check the model

A constant acceleration means equal changes in velocity in equal time intervals. Use separate horizontal and vertical equations only when the motion has been resolved into independent components.

Worked example from local Question Bank row 22687

A glider accelerates uniformly from rest to $27.0\,\mathrm{m\,s^{-1}}$ in $11.0\,\mathrm{s}$. Use $s=\frac{u+v}{2}t$:

$$s=\frac{0+27.0}{2}\times 11.0=148.5\,\mathrm{m}\approx149\,\mathrm{m}$$

The result is the launch-run displacement; the constant-acceleration assumption is essential.

Non-uniform acceleration

Acceleration is non-uniform when its magnitude or direction changes. The velocity–time graph then has a changing gradient, and a single SUVAT value cannot describe the entire interval.

Model versus reality

A constant-acceleration model can be useful over a limited interval even when real forces vary. State the approximation and identify the neglected force or changing condition.

Horizontal motion

There is no horizontal acceleration in the ideal model, so $v_x=u_x$ and $x=u_xt$. Use the horizontal displacement to find time or horizontal speed.

Vertical motion

Use one-dimensional constant-acceleration equations vertically, usually with $a_y=-g$ if upward is positive. The horizontal and vertical equations share the same time $t$.

Worked example from local Question Bank row 31723

A tennis ball travels $11.9\,\mathrm{m}$ horizontally after launch at $64.0\,\mathrm{m\,s^{-1}}$ and $7^\circ$ to the horizontal.

$$u_x=64.0\cos7^\circ=63.52\,\mathrm{m\,s^{-1}}$$
$$t=\frac{x}{u_x}=\frac{11.9}{63.52}=0.187\,\mathrm{s}$$

The same $0.187\,\mathrm{s}$ must then be used in the vertical equation.

Drag opposes instantaneous velocity

Fluid resistance acts opposite the projectile's velocity and usually grows with speed. Its direction changes through the flight, so the resultant acceleration is not the constant downward $g$ of the ideal model.

Use the force direction

Before the peak, drag has horizontal and downward components; after the peak, it has horizontal and upward components. Therefore acceleration is not determined by velocity alone and changes continuously.

Terminal-speed condition

For vertical descent, terminal speed is reached when upward drag (and any buoyancy included in the model) balances weight. The object then continues at constant downward velocity.