Coordinates varying at constant rates | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

r (t) = r_{0} + t v

Initial position plus elapsed time times constant velocity. A collision requires equal positions at one common admissible time.

Constant coordinate rates give a straight path

With constant velocity, each coordinate changes by a fixed amount per unit time. The initial position gives the constant terms; the velocity gives the coefficients of time.

A positive component means that coordinate increases; a negative component means it decreases. Restrict time to the interval for which the model applies.

\begin{aligned} x (t) & = x_{0} + v_{x} t \\ y (t) & = y_{0} + v_{y} t \end{aligned}

Separate speed from displacement

Speed is the magnitude of velocity, so it is nonnegative. Add the squared perpendicular components before taking a square root; adding signed components does not give speed.

Displacement over an interval is velocity multiplied by elapsed time. For constant velocity, distance travelled is speed multiplied by elapsed time.

\begin{aligned} ∣ v ∣ & = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ Δ r & = (t_{2} - t_{1}) v \end{aligned}

A collision needs one common time

For a collision, both objects must have the same coordinates at the same time. Equate their positions using a common clock, and check every coordinate at the candidate time.

Separate parameters can locate a geometric intersection of paths. Compare the arrival times before concluding that the objects collide. Check that each time is in the stated interval.

r_{A} (t) = r_{B} (t)

The coincident straight tracks for A=(1+2t,3-t) and B=(7-t,t/2), shown with solid and dashed traces and opposing direction arrows. Starting positions and their common position (5,1) at t=2 seconds are marked.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find speed and displacement

A particle has position $r (t) = (2 + 3 t, 1 - 4 t)$ metres, with $t$ in seconds. Find its velocity, speed, displacement and distance travelled from $t = 0$ to $t = 5$ .

Worked solution

1
Read the coordinate rates.
$\begin{aligned} r_{0} & = (2, 1) m \\ v & = (3, - 4) m/s \end{aligned}$
The horizontal coordinate increases by $3$ metres each second and the vertical coordinate decreases by $4$ metres each second.
2
Take the magnitude to find speed.
$\begin{aligned} ∣ v ∣ & = \sqrt{3^{2} + {(- 4)}^{2}} \\ = 5 m/s \end{aligned}$
The negative vertical component describes direction; its square contributes positively to speed.
3
Use the elapsed time for both travel quantities.
$\begin{aligned} Δ r & = 5 (\begin{array}{cccccccccc} 3 \\ - 4 \end{array}) \\ = (\begin{array}{cccccccccc} 15 \\ - 20 \end{array}) m \\ d & = 5 \times 5 \\ = 25 m \end{aligned}$
Displacement is a vector. Distance is a nonnegative scalar, and the straight path makes it equal to the displacement magnitude.

Answer

Velocity $(3, - 4)$ m/s; speed $5$ m/s; displacement $(15, - 20)$ m; distance travelled $25$ m.

EXAMPLE 02

Verify a collision in both coordinates

Two particles have positions $r_{A} (t) = (1 + 2 t, 3 - t)$ and $r_{B} (t) = (7 - t, \frac{1}{2} t)$ metres for $t \geq 0$ seconds. Do they collide?

Worked solution

1
Equate the horizontal coordinates.
$\begin{aligned} 1 + 2 t & = 7 - t \\ 3 t & = 6 \\ t & = 2 \end{aligned}$
Any collision must have this time because the horizontal coordinates must match.
2
Check the vertical coordinates at that same time.
$\begin{aligned} y_{A} (2) & = 3 - 2 = 1 \\ y_{B} (2) & = \frac{1}{2} (2) = 1 \end{aligned}$
Both coordinates match at the same admissible time, not merely at two separate times.
3
Find the common position.
$\begin{aligned} r_{A} (2) & = r_{B} (2) \\ = (5, 1) m \end{aligned}$
Substitution into both position rules confirms the collision point.

Answer

They collide at $t = 2$ seconds, at $(5, 1)$ metres.

EXAMPLE 03

Distinguish crossing tracks from a collision

Two carts have positions $r_{A} (t) = (t, 0)$ and $r_{B} (t) = (2, t - 1)$ metres for $t \geq 0$ seconds. Their paths cross at $(2, 0)$ . Do they collide?

Worked solution

1
Find when the first cart reaches the crossing.
$r_{A} (2) = (2, 0)$
The first cart moves along the horizontal axis and reaches the crossing after $2$ seconds.
2
Find when the second cart reaches the crossing.
$t - 1 = 0 ⟹ t = 1$
The second cart has constant horizontal coordinate $2$ and reaches vertical coordinate $0$ after $1$ second.
3
Check the same-time requirement directly.
$\begin{aligned} r_{B} (2) & = (2, 1) \\ \neq (2, 0) \end{aligned}$
Matching horizontal coordinates forces $t = 2$ , but the vertical coordinates differ then. There is no other candidate time.

Answer

No collision. Cart $B$ reaches $(2, 0)$ at $1$ second and cart $A$ reaches it at $2$ seconds.

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Calculate velocity and speed

For $r (t) = (4 - 2 t, 7 + 5 t)$ metres with $t$ in seconds, find velocity and speed.

Show answer & reasoning

$v = (- 2, 5)$ m/s and speed $\sqrt{29}$ m/s.

The coefficients of $t$ are the coordinate rates. Speed is $\sqrt{{(- 2)}^{2} + 5^{2}}$ , not the signed sum $- 2 + 5$ .

CHECK 02

Interpret different arrival times

Cart $A$ has position $(2 t, 0)$ and cart $B$ has position $(6, t - 2)$ metres for $t \geq 0$ seconds. Both tracks contain $(6, 0)$ . Does this prove a collision?

Show answer & reasoning

No. Cart $A$ arrives at $t = 3$ seconds and cart $B$ at $t = 2$ seconds.

At the only candidate common time $t = 3$ , cart $B$ is at $(6, 1)$ . A geometric crossing is insufficient without a common arrival time.

CHECK 03

Apply a meeting model

Two automated vehicles use the same clock. Their positions are $(2 t, 1 + t)$ and $(12 - t, 5)$ metres for $0 \leq t \leq 6$ seconds. Decide whether they meet and give the meeting position.

Show answer & reasoning

They meet at $t = 4$ seconds at $(8, 5)$ metres.

The horizontal equation $2 t = 12 - t$ gives $t = 4$ . The vertical coordinates are then $1 + 4 = 5$ and $5$ , and the time is within the model interval.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Write position from initial $(3, 2)$ m and velocity $(4, - 1)$ m/s.
2 marks
Find speed for $(5 + 8 t, - 2 + 6 t)$ m.
2 marks
A is at $(t, 2 t)$ and B at $(4 - t, 4)$ . Find collision time.
2 marks
A is at $(t, t)$ and B at $(2 - t, 3 - t)$ . Do they collide?
2 marks
A particle moves from $(1, 1, 2)$ to $(7, - 3, 2)$ in $2$ s. Find constant velocity.
2 marks
Find distance travelled by $(1 + 2 t, 3 + t)$ on $0 \leq t \leq 4$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Write position from initial $(3, 2)$ m and velocity $(4, - 1)$ m/s.
2 marks
Use initial position plus $t$ times velocity.
Check at $t = 0$ .
$(3 + 4 t, 2 - t)$ m.
Find speed for $(5 + 8 t, - 2 + 6 t)$ m.
2 marks
Velocity is $(8, 6)$ .
Speed is $\sqrt{64 + 36}$ .
$10$ m/s.
A is at $(t, 2 t)$ and B at $(4 - t, 4)$ . Find collision time.
2 marks
Equate $x$ to get $t = 2$ .
Both $y$ coordinates then equal $4$ .
$t = 2$ , at $(2, 4)$ .
A is at $(t, t)$ and B at $(2 - t, 3 - t)$ . Do they collide?
2 marks
Equal $x$ requires $t = 1$ .
At $t = 1$ , their $y$ values are $1$ and $2$ .
No.
A particle moves from $(1, 1, 2)$ to $(7, - 3, 2)$ in $2$ s. Find constant velocity.
2 marks
Displacement is $(6, - 4, 0)$ .
Divide by $2$ .
$(3, - 2, 0)$ units/s.
Find distance travelled by $(1 + 2 t, 3 + t)$ on $0 \leq t \leq 4$ .
2 marks
Constant speed is $\sqrt{5}$ .
Multiply by elapsed time $4$ .
$4 \sqrt{5}$ units.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics school Vector calculus: polynomial and trigonometric parameterisations, related coordinate rates, Bézier paths, componentwise differentiation and integration, motion, tangent directions, circular motion and arc length. Reasoning and proofs use these school topics; vector-field calculus is outside this chapter.

Official curriculum source ↗

In this lesson

Understand the method.

Constant coordinate rates give a straight path

Separate speed from displacement

A collision needs one common time

See the method in action.

Find speed and displacement

Read the coordinate rates.

Take the magnitude to find speed.

Use the elapsed time for both travel quantities.

Verify a collision in both coordinates

Equate the horizontal coordinates.

Check the vertical coordinates at that same time.

Find the common position.

Distinguish crossing tracks from a collision

Find when the first cart reaches the crossing.

Find when the second cart reaches the crossing.

Check the same-time requirement directly.

Try it yourself.

Calculate velocity and speed

Interpret different arrival times

Apply a meeting model

Keep practising