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:

• IMATH_8
• IMATH_9
• IMATH_10
• IMATH_11
• IMATH_12

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.

Worked example

A ball is thrown vertically upward at $u = 20$ m s$^{-1}$. (a) Maximum height? (b) Time to reach max height? (c) Time of flight back to starting position?

Take up as positive. $a = -g = -9.8$ m s$^{-2}$.

(a) At max height $v = 0$. Use $v^2 = u^2 + 2as$: $0 = 400 + 2(-9.8) s$, so $s = 400/19.6 \approx 20.4$ m.

(b) Use $v = u + at$: $0 = 20 - 9.8 t$, so $t = 20/9.8 \approx 2.04$ s.

(c) Time of flight (back to ground) is twice the time to max height (by symmetry): $\approx 4.08$ s.

Common errors

Confusing distance and displacement. Distance is the total path length; displacement is the net change in position. They differ when the object reverses direction.

Sign errors in free fall. Set up a positive direction and stick to it. Gravity is opposite the upward direction.

Wrong suvat equation choice. Identify your knowns and choose the equation that contains them plus one unknown.

Graph misreading. On a velocity-time graph, the slope is acceleration (not velocity). The area is displacement.

In one sentence

One-dimensional kinematics describes motion using displacement, velocity and acceleration (with sign indicating direction); for constant acceleration the suvat equations ($v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$, $s = \frac{1}{2}(u+v)t$) allow you to solve for any unknown given four of the five variables; graphical analysis uses slope (velocity from position-time, acceleration from velocity-time) and area (displacement from velocity-time, change in velocity from acceleration-time).