Motion in a Straight Line

Displacement and velocity

A particle moves along a straight line. Its displacement s is its position measured from a fixed origin on the line, positive on one side and negative on the other, as a function of the time t.

Its velocity is the rate of change of displacement, $v=\frac{ds}{dt}$v = ds/dt: the gradient of the graph of s against t. If $s=t^2$s = t² meters after t seconds, then v = 2t meters per second. At t = 3 the particle is at s = 9 m and moving at 6 meters per second, and the tangent to the graph there has gradient 6.

Acceleration, and being at rest

Acceleration is the rate of change of velocity, $a=\frac{dv}{dt}$a = dv/dt, the second derivative of s. For v = 2t it is a = 2 meters per second per second: the velocity grows by 2 meters per second every second.

A particle is momentarily at rest when v = 0. That says nothing about a. A ball thrown up with $s=20t-5t^2$s = 20t − 5t² has v = 20 − 10t and a = −10. At t = 2, v = 0 and the ball is at its highest point, 40 − 20 = 20 meters up; its acceleration is still −10, which is why it does not stay there.

The ball $s=20t-5t^2$s = 20t − 5t² on one clock: its height, its velocity 20 − 10t and its acceleration −10. Drag the time to t = 2: the velocity graph crosses 0 at the top of the flight, and the acceleration is −10 throughout.

Velocity has a sign; speed does not

Velocity carries a sign for direction. Positive v means s is increasing, so the particle moves in the positive direction; negative v means it moves the other way. The ball has v = 10 at t = 1, rising at 10 meters per second, and v = −10 at t = 3, falling at 10 meters per second.

Speed is the size of the velocity, speed = |v|, and is never negative. At t = 1 and t = 3 the ball has the same speed, 10 meters per second, and opposite velocities.

A particle that turns back

A particle starts at the origin, and after t seconds its displacement is $s=t^3-6t^2+9t$s = t³ − 6t² + 9t meters, for $t\ge 0$t ≥ 0. Then $v=\frac{ds}{dt}=3t^2-12t+9=3\left(t-1\right)\left(t-3\right)$v = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3) meters per second, and $a=\frac{dv}{dt}=6t-12$a = dv/dt = 6t − 12 meters per second per second.

At t = 0 the particle is moving at 9 meters per second, with a = −12. It is at rest when 3(t − 1)(t − 3) = 0, at t = 1 and at t = 3. At t = 1 it is at s = 1 − 6 + 9 = 4 m, and at t = 3 at s = 27 − 54 + 27 = 0 m, back at the origin.

Between those times the velocity has one sign on each interval. For $0\le t<1$0 ≤ t < 1, v is positive (3.75 at t = 0.5): the particle moves forward. For 1 < t < 3, v is negative (−3 at t = 2): it moves back. For t > 3, v is positive again (3.75 at t = 3.5): forward once more.

The gold curve $s=t^3-6t^2+9t$s = t³ − 6t² + 9t. Its gradient, the velocity, is 0 at (1, 4) and at (3, 0), where the particle turns; between them the graph falls and the particle moves back toward the origin.

Speeding up and slowing down

A particle speeds up when its velocity and acceleration have the same sign, and slows down when they have opposite signs. Negative acceleration alone does not mean slowing down: it pushes in the negative direction, which slows a particle moving forward and speeds up one moving back.

Here a = 6t − 12 is negative for t < 2 and positive for t > 2. So for 0 < t < 1, v is positive and a negative: slowing down, from 9 meters per second to rest. For 1 < t < 2, both are negative: speeding up, backwards. For 2 < t < 3, v is negative and a positive: slowing down, to rest at the origin. For t > 3, both are positive: speeding up.

At t = 2, a = 0 and v = 12 − 24 + 9 = −3. The backward speed stops growing there, so 3 meters per second is the fastest the particle moves backwards.

The gold curve is the velocity $v=3t^2-12t+9$v = 3t² − 12t + 9, below the axis between t = 1 and t = 3, lowest at (2, −3). The dashed curve is the speed |v|, folded above the axis.

Distance and displacement

Over the first 4 seconds the displacement is s(4) − s(0) = (64 − 96 + 36) − 0 = 4 m: the particle ends 4 m from where it began.

The distance traveled is more, because the particle turned back twice. Split the time where v changes sign. From t = 0 to t = 1 it goes from 0 to 4, which is 4 m. From t = 1 to t = 3 it goes from 4 back to 0, another 4 m. From t = 3 to t = 4 it goes from 0 to 4, another 4 m. The distance is 4 + 4 + 4 = 12 m.

The ball is the same. From t = 0 to t = 4 it rises 20 m and falls 20 m, a distance of 40 m, while its displacement is 80 − 80 = 0.

The usual mistakes

Differentiating $s=3t^2$s = 3t² as 3t, or as $6t^2$6t². The power comes down and drops by one: v = 6t.

Giving v as the acceleration. For v = 4t the acceleration is the gradient of that line, 4.

Reading v = 0 as a = 0. The ball at its top has v = 0 and a = −10.

Taking negative acceleration to mean slowing down. Between t = 1 and t = 2 the particle has a < 0 and is speeding up, because it is moving backwards.

Giving the displacement as the distance traveled when the particle turns back: 4 m against 12 m over the first 4 seconds.

Two average-speed cameras

In the application below, a van's distance past a camera is a function of time. Its derivative is the van's speed at each instant, and setting that derivative equal to the average speed between two cameras finds the moment the van was traveling at exactly the average.