Motion That Starts Later

One rule, a delayed clock

A ball is thrown at t = 0 from the origin, and t seconds later it is at $r_1\left(t\right)=\left(2t,20t-5t^2\right)$r₁(t) = (2t, 20t − 5t²) meters. A second ball is thrown from the same point, in the same way, 2 seconds later. Both balls follow the same rule; ball 2 runs it on a clock that started 2 seconds late.

So measure both balls by one clock, t seconds after ball 1 was thrown. At time t, ball 2 has been in the air for t − 2 seconds, so it is where ball 1 was after t − 2 seconds: $r_2\left(t\right)=r_1\left(t-2\right)$r₂(t) = r₁(t − 2).

Put t − 2 in place of t in each coordinate: $r_2\left(t\right)=\left(2\left(t-2\right),20\left(t-2\right)-5{\left(t-2\right)}^2\right)$r₂(t) = (2(t − 2), 20(t − 2) − 5(t − 2)²). Expanded, the first coordinate is 2t − 4, and the second is $20t-40-5\left(t^2-4t+4\right)=-5t^2+40t-60$20t − 40 − 5(t² − 4t + 4) = −5t² + 40t − 60.

When the rule holds

Ball 2 is thrown at t = 2, so $r_2$r₂ describes it only from t = 2 on. Before that the formula still returns numbers: at t = 0 it gives (−4, −60), a point ball 2 never visits. Ball 1 lands when $20t-5t^2=0$20t − 5t² = 0, at t = 4, so ball 2 lands 4 seconds after its own throw, at t = 6. The rule for ball 2 holds for $2\le t\le 6$2 ≤ t ≤ 6.

The two balls at the same moment

At t = 2, ball 1 is at (4, 20), the top of its flight, and ball 2 is at (0, 0), just thrown. At t = 3, ball 1 is at (6, 15), coming down, and ball 2 is at (2, 15), going up: the same height, 4 meters apart. At t = 4, ball 1 is at (8, 0), landing, and ball 2 is at (4, 20), at the top.

In each case ball 2 is where ball 1 was 2 seconds earlier. Check with the expanded rule: at t = 4, 2 × 4 − 4 = 4 and −80 + 160 − 60 = 20.

The first coordinates are 2t and 2t − 4, which differ by 4 at every t. The two balls travel the same path and never meet: ball 2 reaches each point 2 seconds after ball 1 has left it.

The gold curve $y=10x-1.25x^2$y = 10x − 1.25x² is the path of both balls, with the vertical scale squeezed. At t = 3, ball 2 is at (2, 15) going up and ball 1 is at (6, 15) coming down. Ball 2 reaches the top, (4, 20), at t = 4, two seconds after ball 1.

Later subtracts, earlier adds

An object that starts k seconds later than the clock has been moving for t − k seconds at time t, so its position is r(t − k). An object that started k seconds earlier has been moving for t + k seconds, so its position is r(t + k).

The sign is the one to watch. A later start means less time in motion, and less time means a smaller number inside the rule. It is the same shift as in Graph Transformations, where y = f(x − k) moves a graph k to the right: here the graph of each coordinate against t moves k seconds later.

The same path and the same speeds

Differentiate $r_2\left(t\right)=r_1\left(t-2\right)$r₂(t) = r₁(t − 2). By the chain rule, each coordinate is multiplied by the derivative of t − 2, which is 1, so $v_2\left(t\right)=v_1\left(t-2\right)$v₂(t) = v₁(t − 2). Ball 2 has, at each moment, the velocity ball 1 had 2 seconds earlier.

Here $v_1\left(t\right)=\left(2,20-10t\right)$v₁(t) = (2, 20 − 10t), so $v_2\left(t\right)=\left(2,20-10\left(t-2\right)\right)=\left(2,40-10t\right)$v₂(t) = (2, 20 − 10(t − 2)) = (2, 40 − 10t). Differentiating the expanded rule gives the same: the derivative of $-5t^2+40t-60$−5t² + 40t − 60 is 40 − 10t. Ball 2's upward velocity is 0 at t = 4, its top, 2 seconds after ball 1's at t = 2.

Meeting, with a later start

Boat A leaves a harbor at the origin at t = 0, where t is in hours, with velocity (4, 3) kilometers per hour, so rA = (4t, 3t). Boat B leaves a buoy at (20, −3) at t = 1 with velocity (−4, 6) kilometers per hour. At time t it has been moving for t − 1 hours, so rB = (20 − 4(t − 1), −3 + 6(t − 1)) for $t\ge 1$t ≥ 1.

The boats meet if both coordinates agree at one value of t. First coordinates: 4t = 20 − 4(t − 1) = 24 − 4t, so 8t = 24 and t = 3. Second coordinates at t = 3: boat A is at 3 × 3 = 9 and boat B at −3 + 6 × 2 = 9. They agree, and t = 3 is after B sets off, so the boats meet at (12, 9) at t = 3.

That last check matters. Boat C leaves (16, 12) at t = 5 with velocity (2, 1.5), so rC = (16 + 2(t − 5), 12 + 1.5(t − 5)) = (2t + 6, 1.5t + 4.5) for $t\ge 5$t ≥ 5. Setting 4t = 2t + 6 gives t = 3, and the second coordinates agree there: 9 and 4.5 + 4.5 = 9. But t = 3 is before C sets off, when C is still at its buoy. Boat A passes (16, 12) at t = 4, an hour before C leaves it, and from t = 5 on A is ahead and faster. They never meet.

Boat A from the origin, setting off at t = 0, and boat B from (20, −3), setting off at t = 1. A takes 3 hours to reach (12, 9) and B takes 2, so both arrive at t = 3.

The usual mistakes

Writing r(t + k) for a later start. That runs the object ahead of the clock instead of behind it.

Writing r(t) − k. That moves the whole path k units in space; the object travels the same path, later.

Giving A's position for B's. At t = 3 ball 1 is at (6, 15); ball 2 has been going for only 1 second and is at (2, 15).

Using the rule before the object starts. A solution with t < k describes a moment when the object has not set off, so it is not a meeting.

Matching one coordinate and stopping. Both coordinates must agree at the same t.

A launch sent after a boat

In the first application below, the launch sets off twenty minutes after the cruiser, so its position is written in m − 20, and the meeting time it finds, m = 40, is checked against $m\ge 20$m ≥ 20.

A course that changes partway

In the second, a robot vacuum turns after 20 seconds. Its second leg starts at the end of the first, and the time to use on that leg is the time since the turn, m − 20.