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

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

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

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

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

Now you

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

An object has velocity (6, 8). How far does it travel each second?

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

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

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

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

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

Now you

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

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

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

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

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

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

Now you

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

|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

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

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

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

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

Now you

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

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

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

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

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

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

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

Now you

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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