Kinematics of motion in one dimension: displacement, velocity, acceleration, the equations of uniformly accelerated motion (suvat), and graphical analysis

What this dot point is asking

VCAA wants you to describe motion in one dimension using displacement, velocity and acceleration, apply the suvat (constant-acceleration) equations to motion problems, and interpret motion graphically.

Definitions

Position $x$

Location of an object along the chosen axis. Vector quantity in higher dimensions but scalar (positive or negative) in one dimension.

Displacement $\Delta x$

Change in position. Vector. Different from distance (a scalar).

Velocity $v$

Rate of change of displacement: $v = \Delta x / \Delta t$. Vector. Sign indicates direction.

Speed $|v|$

Magnitude of velocity. Always non-negative.

Acceleration $a$

Rate of change of velocity: $a = \Delta v / \Delta t$.

Average vs instantaneous

Average is over a finite time interval. Instantaneous is the limit as $\Delta t \to 0$ (the derivative).

SUVAT equations (uniformly accelerated motion)

For constant acceleration:

• $v = u + at$
• $s = ut + \frac{1}{2} a t^2$
• $v^2 = u^2 + 2 a s$
• $s = \frac{1}{2}(u + v) t$
• $s = vt - \frac{1}{2} a t^2$

Where $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $s$ = displacement, $t$ = time.

Choose the equation that contains the four known variables (and the one unknown).

Free fall

An object in free fall near Earth's surface has $a = g \approx 9.8$ m s$^{-2}$ directed downward. (Ignoring air resistance.)

Apply suvat with $a = -g$ (taking up as positive) or $a = +g$ (taking down as positive). Sign consistency is essential.

Graphical analysis

Position-time graph

Slope = instantaneous velocity at that moment.

Velocity-time graph

Slope = instantaneous acceleration. Area under the curve = displacement.

Acceleration-time graph

Area under the curve = change in velocity.

A horizontal line on a velocity-time graph means constant velocity (zero acceleration). A straight line with constant positive slope means constant positive acceleration.

Examples in context

Example 1. Tullamarine Airport runway take-off roll. A Qantas A330 ($230\,000$ kg) accelerates from rest along Tullamarine's runway 16/34 (length $3657$ m) and lifts off at $80$ m s$^{-1}$. Assuming uniform acceleration, $v^2 = u^2 + 2as$ gives $a = v^2/(2s) = 6400/(2 \times 1600) = 2.0$ m s$^{-2}$ over the $1600$ m to rotation, with $t = v/a = 40$ s. The remaining $2000$ m of runway is safety overrun. Pilots cross-check this acceleration with V$_1$ (decision speed) and V$_R$ (rotation speed) markers; if acceleration is less than expected, the pilot aborts before V$_1$.

Example 2. Free fall test of an FMG drill core at Mt Stromlo. A geophysics lab at Mt Stromlo drops a $0.5$ m rock core into a vacuum tube to time-of-flight test gravimeter calibration. Free fall from rest gives $s = \tfrac{1}{2}gt^2$. For $s = 1.0$ m and $g = 9.80$ m s$^{-2}$, $t = \sqrt{2s/g} = \sqrt{2/9.80} = 0.452$ s and final velocity $v = gt = 4.43$ m s$^{-1}$. Photogate timing to four decimal places lets the lab resolve $g$ to $\pm 0.001$ m s$^{-2}$, sufficient to detect Earth-tide variations and to calibrate field gravimeters used in iron-ore exploration.

Try this

Q1. Define uniform acceleration and state the units of acceleration. [2 marks]

• Cue. Constant rate of change of velocity with time; m s$^{-2}$.

Q2. A car starting from rest reaches $30$ m s$^{-1}$ over $150$ m. Calculate (a) the acceleration, and (b) the time taken. [4 marks]

• Cue. (a) $v^2 = 2as$, $a = 900/300 = 3.0$ m s$^{-2}$. (b) $t = v/a = 10$ s.

Q3. Refer to the Tullamarine A330 example. (a) Calculate the rotation acceleration. (b) Determine the time to lift-off. (c) Explain why airlines require excess runway beyond rotation distance. [2+2+2 marks]

• Cue. (a) $2.0$ m s$^{-2}$. (b) $40$ s. (c) Engine-out safety: aircraft must accelerate, abort and stop within runway length even if one engine fails.