VCE Physics (Year 12) · Motion and Forces · 18m read

Kinematics — Motion in a Straight Line

Ever wondered how to predict exactly where a moving car will be in 3 seconds, or how fast it needs to go to stop in time? That's kinematics—the toolkit for describing motion using just distance, speed, and time. You'll learn five equations (often called SUVAT) that connect these quantities, plus how to read and draw motion graphs that show what's really happening. Mastering this foundation makes every other physics topic—from projectiles to collisions—click into place, and it's essential for VCE exams and real safety calculations in engineering.

Learning objective: Use the five kinematic equations and motion graphs to solve straight-line motion problems involving constant acceleration.

By the end of this lesson you can:

I can distinguish scalar distance/speed from vector displacement/velocity/acceleration.
I select the correct SUVAT equation based on known and unknown variables.
I correctly apply sign convention throughout multi-step problems.
I extract acceleration from a v–t gradient and displacement from a v–t area.
I solve two-phase problems (e.g. acceleration then deceleration).

Kinematics is a car's trip computer. It logs displacement, velocity and acceleration — and the SUVAT equations predict the rest of the trip.

Scalars vs Vectors — the essential distinction

Physics splits physical quantities into two categories that behave very differently:

Scalars have magnitude only.
- *Distance* $d$: total path length travelled (always $\geq 0$).
- *Speed* $v_{sp}$: magnitude of velocity ($|v|$), always $\geq 0$.
- Time $t$, mass $m$, temperature $T$.

Vectors have both magnitude and direction.
- *Displacement* $s$ (or $\vec{s}$, or $\Delta x$): straight-line distance from start to finish, *with* direction.
- *Velocity* $v$ (or $\vec{v}$): rate of change of *displacement*.
- *Acceleration* $a$ (or $\vec{a}$): rate of change of *velocity*.

> Key insight: A car that drives 30 km north and then 30 km south has travelled a *distance* of 60 km but has a *displacement* of 0 km. These are physically different quantities.

### Sign convention

In one-dimensional problems you assign one direction as positive (usually right, up, or the initial direction of motion). The opposite direction is then negative. Once you choose, you must stick with it throughout the entire problem. Most VCAA errors come from inconsistent sign convention.

Example: A ball thrown upward at $u = +15 \text{m/s}$ with $g = 9.8 \text{m/s}^2$ downward means $a = -9.8 \text{m/s}^2$ (taking up as positive).

The five SUVAT equations — derivation and use

SUVAT equations apply whenever acceleration $a$ is constant (uniform). The five variables are:

| Symbol | Quantity | SI unit |
|--------|----------|---------|
| $s$ | displacement | m |
| $u$ | initial velocity | m/s |
| $v$ | final velocity | m/s |
| $a$ | acceleration | m/s² |
| $t$ | time | s |

Each equation omits one variable — that's how you choose which to use:

$$v = u + at \quad \text{(omits } s\text{)}$$

$$s = ut + \tfrac{1}{2}at^2 \quad \text{(omits } v\text{)}$$

$$v^2 = u^2 + 2as \quad \text{(omits } t\text{)}$$

$$s = \tfrac{1}{2}(u + v)t \quad \text{(omits } a\text{)}$$

$$s = vt - \tfrac{1}{2}at^2 \quad \text{(omits } u\text{)}$$

### How the equations are derived

Start from the definition of constant acceleration:

$$a = \frac{v - u}{t} \implies v = u + at \quad \text{...(1)}$$

Average velocity (constant $a$ means linear $v$–$t$):

$$\bar{v} = \frac{u + v}{2} \implies s = \frac{u+v}{2} \cdot t \quad \text{...(4)}$$

Substitute (1) into (4):

$$s = \frac{u + u + at}{2} \cdot t = ut + \tfrac{1}{2}at^2 \quad \text{...(2)}$$

Eliminate $t$ between (1) and (4):

$$t = \frac{v - u}{a}, \quad s = \frac{v+u}{2} \cdot \frac{v-u}{a} = \frac{v^2 - u^2}{2a}$$

$$\therefore v^2 = u^2 + 2as \quad \text{...(3)}$$

Understanding the derivation means you can reconstruct any forgotten equation from first principles in an exam.

SUVAT Formula Sheet

No displacement: v = u + at (Use when $s$ is not involved)
No final velocity: s = ut + \tfrac{1}{2}at^2 (Use when $v$ is not needed)
No time: v^2 = u^2 + 2as (Most powerful — no $t$ required)
No acceleration: s = \tfrac{1}{2}(u+v)t (Use when $a$ not given)
No initial velocity: s = vt - \tfrac{1}{2}at^2 (Rare — when $u$ unknown)
Free fall only: g = 9.8 \ \text{m/s}^2 \ \text{(downward)} (VCAA standard value)

Motion graphs — position, velocity and acceleration

Motion graphs are tested in almost every VCAA exam. You need to read gradients and areas fluently.

### Position–time graph ($x$–$t$)

Gradient at any point = instantaneous velocity $v = \dfrac{\Delta x}{\Delta t}$
Straight line → constant velocity (zero acceleration)
Curve → changing velocity (non-zero acceleration)
Parabola → constant acceleration (from SUVAT equation 2)
Object at rest: horizontal line; returning to origin: negative gradient

### Velocity–time graph ($v$–$t$)

Gradient = acceleration $a = \dfrac{\Delta v}{\Delta t}$
Area under curve (above/below axis) = displacement $s$
Constant acceleration → straight line on $v$–$t$
Area *above* the $t$-axis = positive displacement; area *below* = negative displacement

### Acceleration–time graph ($a$–$t$)

Area = change in velocity $\Delta v$
Horizontal line (constant $a$) is the most common type in VCAA

### Practical extraction skills

To find displacement from a $v$–$t$ graph with a trapezoidal shape:
$$s = \text{area of trapezoid} = \frac{1}{2}(v_1 + v_2) \cdot t$$

To find acceleration from two points on a $v$–$t$ graph:
$$a = \frac{v_2 - v_1}{t_2 - t_1}$$

A typical v–t graph: speed up, cruise, slow down: A typical VCAA-style v–t graph: Area = displacement; Gradient = acceleration. The steeper the line, the larger the acceleration. Here: accelerate at 3 m/s² for 4 s, cruise at 12 m/s for 6 s, then decelerate at 3 m/s². Displacement = ½(4)(12) + 6(12) + ½(4)(12) = 120 m.

SUVAT problem-solving protocol

1. Set direction: Choose positive direction before writing anything. Label it on your diagram.
2. List all 5 variables: Write $s =$ , $u =$ , $v =$ , $a =$ , $t =$. Fill in knowns; mark unknown with ?.
3. Count your knowns: You need 3 known values to use any SUVAT equation and solve for 1 unknown.
4. Pick the equation: Find the SUVAT equation that contains your 3 knowns + the unknown. The missing variable tells you which formula to use.
5. Substitute carefully: Include signs. $u = -4 \text{m/s}$ means the object initially moves in the negative direction.
6. Check reasonableness: Does the sign make sense? Is the magnitude plausible? Negatives for displacement mean the object moved backward.

Worked example (Easy)

A train starts from rest and accelerates uniformly at $1.5 \ \text{m/s}^2$ for $10 \ \text{s}$. Find its final velocity and distance travelled.

List knowns and unknowns. u = 0 (starts from rest), a = 1.5 m/s², t = 10 s. Find v and s. u = 0, \ a = 1.5 \ \text{m/s}^2, \ t = 10 \ \text{s}
Find final velocity using equation (1). Use $v = u + at$ because we know $u$, $a$, $t$ and want $v$. v = 0 + (1.5)(10) = 15 \ \text{m/s}
Find displacement using equation (2). Use $s = ut + \frac{1}{2}at^2$ because we know $u$, $a$, $t$ and want $s$. s = 0(10) + \tfrac{1}{2}(1.5)(10)^2 = 0 + 75 = 75 \ \text{m}
Verify with equation (4). Cross-check with $s = \frac{1}{2}(u+v)t$ — a second method is always worth checking. s = \tfrac{1}{2}(0 + 15)(10) = 75 \ \text{m} \checkmark

Answer: Final velocity $v = 15 \ \text{m/s}$; displacement $s = 75 \ \text{m}$.

Worked example (Medium)

A car travelling at $20 \ \text{m/s}$ brakes uniformly and stops in $40 \ \text{m}$. Find the deceleration and the braking time.

Set up and list knowns. Taking forward as positive: u = +20 m/s, v = 0 (stops), s = +40 m. We want a and t. u = +20, \ v = 0, \ s = +40 \ \text{m}
Find acceleration using equation (3). Use $v^2 = u^2 + 2as$ — we know $v$, $u$, $s$ and want $a$. This is the no-time equation. 0 = (20)^2 + 2a(40) \implies a = \frac{-400}{80} = -5.0 \ \text{m/s}^2
Interpret the sign. a = −5.0 m/s² confirms deceleration (opposing the positive direction of travel). The magnitude of deceleration is 5.0 m/s².
Find time using equation (1). Now use $v = u + at$ — rearrange for $t$. 0 = 20 + (-5.0)t \implies t = \frac{20}{5.0} = 4.0 \ \text{s}

Answer: Deceleration $a = 5.0 \ \text{m/s}^2$; braking time $t = 4.0 \ \text{s}$.

Worked example (Hard)

A ball is thrown vertically upward at $14.7 \ \text{m/s}$ from the top of a cliff 19.6 m above the ground. Taking up as positive and $g = 9.8 \ \text{m/s}^2$, find: (a) the maximum height above the cliff top, (b) the total time before it hits the ground, (c) its speed on impact.

Set up sign convention. Up = positive. Initial position = cliff top (our reference, s = 0). Ground is 19.6 m below → ground is at s = −19.6 m. u = +14.7 m/s, a = −9.8 m/s². u = +14.7 \ \text{m/s}, \ a = -9.8 \ \text{m/s}^2
(a) Maximum height: v = 0 at the top. At maximum height, velocity = 0. Use $v^2 = u^2 + 2as$. 0 = (14.7)^2 + 2(-9.8)s_{\max} \implies s_{\max} = \frac{216.09}{19.6} = 11.025 \ \text{m} \approx 11.0 \ \text{m above cliff}
(b) Total time to ground: s = −19.6 m. Use $s = ut + \frac{1}{2}at^2$. The displacement when it hits the ground is −19.6 m (ground is below throw point). -19.6 = 14.7t + \tfrac{1}{2}(-9.8)t^2 = 14.7t - 4.9t^2
Rearrange to quadratic. Move everything to one side. 4.9t^2 - 14.7t - 19.6 = 0 \implies t^2 - 3t - 4 = 0 \quad (\div 4.9)
Factorise and solve. Factor the quadratic. (t - 4)(t + 1) = 0 \implies t = 4.0 \ \text{s} \ (\text{reject } t = -1)
(c) Speed at impact. Use $v = u + at$ with t = 4.0 s. v = 14.7 + (-9.8)(4.0) = 14.7 - 39.2 = -24.5 \ \text{m/s}
Interpret final answer. v = −24.5 m/s means the ball is moving downward at impact. Speed (magnitude) = 24.5 m/s.

Answer: (a) $s_{\max} = 11.0 \ \text{m}$ above cliff top. (b) $t = 4.0 \ \text{s}$. (c) Speed at impact $= 24.5 \ \text{m/s}$.

Common mistakes

Using $g = +9.8 \ \text{m/s}^2$ upward Gravity always acts downward. If you take up as positive, then $a = -9.8 \ \text{m/s}^2$. Writing $a = +9.8$ when up is positive is a sign error that corrupts every calculation in the problem. Fix: Explicitly write your sign convention at the start: "Taking up as positive, $a = -9.8 \ \text{m/s}^2$." Then every substitution is automatic.
Confusing displacement $s$ with distance $d$ If a ball goes up 10 m and falls back 10 m to the same height, displacement $s = 0$ but distance $d = 20$ m. SUVAT uses displacement — using distance will give a wrong answer. Fix: Sketch the motion with arrows. Clearly mark *where the object starts*, *where it ends*, and the sign of each phase.
Only one SUVAT equation for a two-phase problem If acceleration changes partway through (e.g. rocket burns, then coasts), the conditions change and SUVAT cannot span both phases as a single equation. Fix: Split at the point where conditions change. Solve Phase 1, use its final state as Phase 2's initial state.
Forgetting to reject the unphysical root in a quadratic Solving $s = ut + \frac{1}{2}at^2$ as a quadratic often gives two roots for $t$. The negative root is mathematically valid but physically meaningless (time cannot be negative). Fix: Write both roots, then explicitly state "reject $t = $ [negative value] since time $\geq 0$."

Why does SUVAT require constant acceleration? And what if it's not constant?

SUVAT equations are derived by integrating constant acceleration. The key mathematical step is:

$$v(t) = \int a \, dt = at + C$$

With $v(0) = u$, we get $C = u$, so $v = u + at$ — only valid when $a$ is truly constant.

### When acceleration varies

If $a = f(t)$ (not constant), you must use calculus:

$$v(t) = u + \int_0^t a(t') \, dt'$$
$$s(t) = \int_0^t v(t') \, dt'$$

For example, if $a(t) = 3t$ (acceleration increasing linearly):

$$v(t) = u + \frac{3t^2}{2}, \qquad s(t) = ut + \frac{t^3}{2}$$

These cannot be found using SUVAT.

### On a $v$–$t$ graph

Constant acceleration = straight line. Variable acceleration = a curve. The area under the curve is still displacement:

$$s = \int_0^T v(t) \, dt$$

At Year 12 you estimate this numerically (counting squares or using trapezoid rule). At university you use analytic integration.

### Practical check

Always ask: *"Is acceleration constant throughout the interval I'm using SUVAT for?"* If the problem says "uniform acceleration" or "constant braking force" — yes. If the object is on a slope with air resistance, friction, or a varying engine force — potentially no.

Practice questions

Which SUVAT equation is best to use when time is unknown?
1. $v = u + at$
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$
4. $s = \frac{1}{2}(u+v)t$
(1 marks)
A stone is dropped from rest from a cliff. After $3.0 \ \text{s}$, its velocity is (take $g = 9.8 \ \text{m/s}^2$):
1. $9.8 \ \text{m/s}$ downward
2. $19.6 \ \text{m/s}$ downward
3. $29.4 \ \text{m/s}$ downward
4. $44.1 \ \text{m/s}$ downward
(2 marks)
A cyclist travelling at $8.0 \ \text{m/s}$ applies brakes uniformly and stops in $16 \ \text{m}$. Find the magnitude of deceleration.
1. $1.0 \ \text{m/s}^2$
2. $2.0 \ \text{m/s}^2$
3. $3.0 \ \text{m/s}^2$
4. $4.0 \ \text{m/s}^2$
(3 marks)
A ball is thrown vertically upward at $19.6 \ \text{m/s}$. Taking up as positive and $g = 9.8 \ \text{m/s}^2$, what is the maximum height reached above the launch point?
1. $9.8 \ \text{m}$
2. $19.6 \ \text{m}$
3. $29.4 \ \text{m}$
4. $39.2 \ \text{m}$
(3 marks)
A car accelerates from $10 \ \text{m/s}$ to $30 \ \text{m/s}$ over $200 \ \text{m}$. It then decelerates uniformly to rest in a further $4.0 \ \text{s}$. Find: (a) the acceleration in the first phase, and (b) the distance covered in the second phase.
(4 marks)
A velocity–time graph for a moving object shows a straight line from $(0, -6)$ to $(4, 10)$. Determine (a) the acceleration, (b) the displacement over 4 s, and (c) the distance travelled over 4 s.
(4 marks)

Summary

SUVAT equations only apply when acceleration a is constant (uniform).
Five variables: $s$ (displacement), $u$ (initial $v$), $v$ (final $v$), $a$ (acceleration), $t$ (time). Need 3 knowns to find 1 unknown.
$v^2 = u^2 + 2as$ is the most useful — no time required.
Sign convention: pick a positive direction, assign negative to the opposite. Apply consistently to $u$, $v$, $a$, and $s$.
On a $v$–$t$ graph: gradient $= a$; area under curve $= s$ (net displacement, not distance).
Two-phase problems: use the end state of phase 1 as the start state of phase 2.
Quadratic roots: always reject physically impossible (negative) time.