Kinematics of one-dimensional motion: VCE Physics Unit 2 Year 11

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.