Motion in Two Dimensions

Stage 19 of 23 Strand 3 of 5 5 lessons

5 illustrated lessons, each teaching the why before the how.

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Position and Velocity as Vectors

One straight line, written r = r0 + vt.

A constant velocity moves a position vector along the straight line r = r0 + vt

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xy

At time zero the object is at r₀ = (1, 2), and has not yet moved.

xyv

Its velocity is v = (3, 2): every second it moves 3 across and 2 up.

xyt = 0t = 1t = 2

Add v again for each second that passes. The positions lie on one straight line.

r = (1, 2) + t(3, 2)one equation per coordinater = (1 + 3t, 2 + 2t)

Written out, each coordinate is its own straight-line equation in t.

Now you

r = (2, 1) + 3(3, 3)

r₀ = (2, 1) and v = (3, 3). Where is the object at t = 3?

r = (3, 2) + 3(2, 3)

r₀ = (3, 2) and v = (2, 3). Where is the object at t = 3?

Meeting Points and Travel Times

Crossing paths is not the same as colliding.

Two paths crossing is not a collision unless both objects are there at the same time

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xyABpaths cross

These two straight paths cross. That is a fact about the routes, not about the objects on them.

rA = (1 + 3t, 2 + 2t)rB = (10 − t, 2 + 3t)both coordinates, at the same instantA collision needs rA = rB at one value of t

For a collision both coordinates must agree, and agree at the same value of t.

rA = rBthe x-coordinates1 + 3t = 10 − t4t = 9, t = 2.25the y-coordinates disagree2 + 2(2.25) = 6.5 and 2 + 3(2.25) = 8.75

Solve one coordinate for t, then test the other. Here the y-coordinates disagree, so they never collide.

rA = (1 + 3t, 2 + 2t), rB = (10 − t, 6 − t)1 + 3t = 10 − t gives t = 2.25the y-coordinates disagree again2 + 2(2.25) = 6.5 and 6 − 2.25 = 3.75

Paths can cross while the objects pass the crossing point hours apart.

Now you

3 + 3t = 21 − 1tgather the t terms4t = 18

A is at 3 + 3t and B at 21 − 1t on the same axis. When are they level?

1 + 3t = 16 − 1tgather the t terms4t = 15

A is at 1 + 3t and B at 16 − 1t on the same axis. When are they level?

The Closest Approach of Two Objects

The squared gap is a quadratic in time.

The gap between two objects is a quadratic in time, so it has one lowest value

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rA = (1 + 3t, 2 + 2t), rB = (10 − t, 6 − t)subtract, coordinate by coordinateAB = rB − rA = (9 − 4t, 4 − 3t)

Subtract one position vector from the other to get the gap between them, as a vector in t.

a quadratic in t, opening upwards

Work with the square of the length: the algebra is easier, and it is smallest at the same moment.

xyclosest

The curve turns once. That turning point is the moment the two are nearest.

the turning point50t − 96 = 0, t = 1.92take the square root at the end
, so |AB| = 2.2

Differentiate, set the derivative to 0 to find t, then take the square root for the distance.

Now you

at the turning point

The smallest value of |AB|² is 9. What is the closest approach?

differentiate

|AB|² = 2t² − 16t + 31. At what time are they closest?

Velocity That Varies with Time

Differentiate the position vector coordinatewise.

Differentiating a position vector coordinate by coordinate gives velocity, then acceleration

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A thrown ball moves steadily across, while a squared term pulls the height back down.

xy

Plot one coordinate against the other and the path appears: a parabola.

differentiate each coordinatev = (2, 5 − 10t)differentiate againa = (0, −10)

The horizontal velocity never changes, and the vertical velocity falls by 10 every second.

a circle of radius 1r(t) = (cos t, sin t)v = (−sin t, cos t)pointing straight back at the centera = (−cos t, −sin t) = −r

Circular motion: the acceleration always points back at the center of the circle.

Now you

differentiate each coordinate

r(t) = (5t, 9t − 3t²). What is v at t = 2?

r(t) = (cos t, sin t)differentiate each coordinatev = (−sin t, cos t)

r(t) = (cos t, sin t). What is the acceleration?

Motion That Starts Later

Replace t by t − k and the path is unchanged.

An object setting off k seconds later has position r(t − k), the same rule on a delayed clock

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thrown at t = 0

One ball is thrown at t = 0, and this rule gives where it is at every later t.

thrown 2 seconds later

Ball 2, thrown 2 seconds later, is at time t where ball 1 was at time t − 2.

ball 1ball 2t = 2(4, 20)(0, 0)t = 3(6, 15)(2, 15)t = 4(8, 0)(4, 20)

At t = 4 ball 2 is at the top, where ball 1 was two seconds earlier.

starts later
this sign is the one to watch
starts earlier

A later start subtracts: the clock inside the rule runs behind the clock outside.

the derivative of t − 2 is 1, so nothing else changes

The speed and the path are unchanged — only the time at which each position is reached moves.

Now you

A: r(t)B is k seconds behind AB: ?

Object B follows A’s path but sets off k seconds later. What is B’s position rule?

?)

B copies A’s path from 1 second later. When is B where A was at t = 2?

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