Accumulate signed velocity and total travel

Try

A particle moves with velocity v(t)=2t−4 metres per second from t=0 to t=3 seconds. It starts moving negatively and later reverses. Predict whether its net displacement can have smaller magnitude than its total distance traveled. Draw a number line with separate arrows for the two stages of motion.

Predict

Set the velocity equal to zero to find the potential reversal time. Then check its sign on both sides; a zero alone would not prove reversal. Predict whether average speed equals the magnitude of average velocity for this trip.

Experiment

The primitive of velocity is t²−4t. Build a ledger using the sign change at t=2.

Time interval	Velocity sign	Signed displacement	Distance
[0,2]	negative inside	−4 m	4 m
[2,3]	positive inside	1 m	1 m
Total	reversal	−3 m	5 m

Sketch the velocity line and label the triangular signed areas. If initial position is 10 metres, final position is 7 metres, not −3 metres. The initial position shifts the whole trajectory but does not change the trip’s displacement or distance. Compare these quantities in the signed-motion investigation.

Understand

When s′(t)=v(t), the fundamental theorem gives s(b)−s(a)=∫ₐᵇv(t)dt. This is displacement, a signed change in position. Total distance is ∫ₐᵇ|v(t)|dt when the velocity model permits that accumulation. Split at velocity sign changes to remove the absolute value correctly. Integrating v and taking an absolute value only afterward generally loses travel canceled by a reversal.

Average velocity divides displacement by elapsed time; average speed divides total distance by elapsed time. Here they are −1 and 5/3 metres per second. The magnitude of average velocity is at most average speed, with equality when motion does not reverse apart from irrelevant zero-velocity intervals. The denominator is elapsed time, not the number of observations.

Acceleration requires two integrations to recover position. If a(t)=6t, then v(t)=3t²+C₁. The condition v(0)=2 gives C₁=2. Integrating again gives s(t)=t³+2t+C₂, and s(0)=−1 gives C₂=−1. Thus s(2)=11. A known acceleration without initial velocity and position does not uniquely specify a trajectory.

Physical interpretation depends on units and the validity of the model. Integrating metres per second against seconds produces metres. If time is recorded in minutes, convert consistently. Do not combine a rate in one time unit with a width in another and assume the result is correct because the arithmetic is simple.

Endpoint observations alone cannot recover a trip. The continuous velocity models v(t)=0 and v(t)=t(1−t) on [0,1] both vanish at their endpoints but have different integrals. A statement that a car was stopped at the beginning and end says nothing by itself about how far it traveled between them. Numerical interpolation and measurements require assumptions beyond the symbolic integral rules.

Apply

Use the practice tasks to distinguish final position, displacement, distance and two kinds of average. For a derivation from acceleration, solve for both constants explicitly. For the sparse-data counterexample, show that both proposed models satisfy the observations before comparing their totals.

Continue

Next accumulate areas of thin cross sections to obtain volume. The same principle—local contribution times thickness—remains, but the local quantity is now an area rather than a velocity.