Coordinates varying at constant rates | Year 12

01 / LEARN

The idea, step by step

Constant rates give a straight path

Constant velocity gives $r (t) = r_{0} + t v$ . The initial vector is the position at t=0; the velocity vector contains the coordinate rates. Speed is its magnitude, not the sum of components.

For $r = (x_{0} + a t, y_{0} + b t)$ , the direction is (a, b). A restricted time interval gives a segment rather than a whole line. Use the same position and time units throughout.

A collision needs one common time

Equate both coordinates using the same t. A value satisfying only the x equation is insufficient. In three dimensions, the z coordinate must match too.

Paths can intersect at different times. Separate parameters may find a geometric intersection, but compare arrival times before claiming a collision.

A common trap

Intersecting paths prove a collision.

Instead: Positions must agree at the same time.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

A has position $(1 + 2 t, 3 - t)$ and B has $(7 - t, t / 2)$ metres. Do they collide for $t \geq 0$ ?

Equate x: $1 + 2 t = 7 - t$ , giving t=2.
At that time both y coordinates equal 1.
Both position coordinates match at the same admissible time.

At t=2 s, at (5, 1) m.

Example 2

Worked example

Find speed and 5-second displacement for $r = (2 + 3 t, 1 - 4 t)$ m.

Velocity is (3, -4).
Its magnitude is 5.
Multiply velocity by 5 for displacement.

Speed 5 m/s; displacement (15, -20) m.

Check your understanding

Find velocity for $(4 - 2 t, 7 + 5 t)$ .

Show answer

$(- 2, 5)$

The t coefficients are coordinate rates.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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 topic 6.3–6.5: polynomial and trigonometric parameterisations, related coordinate rates, motion, tangent directions and arc length. This is school Vector calculus, not vector-field calculus.

Official curriculum source ↗

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