MCV4U Motion: Displacement, Velocity and Acceleration: Notes and Practice

A phone's GPS records only one thing: where you are at each moment. Yet a map app can tell you how fast you are going and whether you are braking. It does this by watching how quickly your position changes, which is exactly what a derivative measures. In this lesson an object moves back and forth along a straight line (a car on a straight road, an elevator, a ball thrown straight up), and its position at time $t$ is a function $s (t)$ . Everything else about the motion comes from differentiating that one function.

Velocity is the instantaneous rate of change of position, so it is the derivative $s^{'} (t)$ . Acceleration is the instantaneous rate of change of velocity, so it is the derivative of velocity, which makes it the second derivative of position. If $s$ is in metres and $t$ in seconds, then $v$ is in metres per second and $a$ is in metres per second per second, written $m/s^{2}$ .

v (t) = s^{'} (t) = \frac{d s}{d t}, a (t) = v^{'} (t) = s^{''} (t) = \frac{d ^{2} s}{d t ^{2}}

Each quantity is the rate of change of the one before it.

Sign	What it means
$s (t) > 0$	The object is on the positive side of the starting reference point (the origin).
$v (t) > 0$	The object is moving in the positive direction (right, or up).
$v (t) < 0$	The object is moving in the negative direction (left, or down).
$v (t) = 0$	The object is momentarily at rest.
$a (t) > 0$	The velocity is increasing (more positive or less negative), which is NOT the same as speeding up.
$a (t) < 0$	The velocity is decreasing (less positive or more negative), which is NOT the same as slowing down.

Velocity has a direction built into its sign. Speed is the size of the velocity, |v(t)|, and is never negative.

Fig 1Position, velocity and acceleration graphs for s(t) = t³ − 6t² + 9t from t = 0 to t = 4, lined up on the same time axis. The object is at rest at t = 1 and t = 3, where v = 0 and the direction changes; the intervals where v and a have the same sign (speeding up), 1 < t < 2 and 3 < t < 4, are shaded.

At rest and changing direction

An object is at rest at the instants when $v (t) = 0$ . It changes direction only if the velocity changes SIGN there, from positive to negative or the other way. A ball thrown straight up stops for an instant at the top and then falls, so its velocity goes from positive to negative and it changes direction. But an object with $v (t) = (t - 2)^{2}$ is at rest at $t = 2$ and then keeps going the same way, because $(t - 2)^{2}$ is never negative. To be sure, check the sign of $v$ on both sides of each zero, just as you did with the first derivative test.

Speeding up or slowing down

Compare the signs of v and a

An object is speeding up when

v

and

a

have the SAME sign, and slowing down when they have OPPOSITE signs. A car reversing out of a driveway has

v < 0

; if it is also accelerating backward (

a < 0

), its speed is growing. A negative acceleration does not mean slowing down on its own: it slows the object only when the object is moving in the positive direction.

Interval	Sign of v = 3(t − 1)(t − 3)	Sign of a = 6t − 12	Motion
$0 < t < 1$	$+$	$-$	Moving forward, slowing down
$1 < t < 2$	$-$	$-$	Moving backward, speeding up
$2 < t < 3$	$-$	$+$	Moving backward, slowing down
$t > 3$	$+$	$+$	Moving forward, speeding up

Sign analysis for s(t) = t³ − 6t² + 9t. The breakpoints are the zeros of v (t = 1, 3) and the zero of a (t = 2).

VideoSal Khan analyzes the same position function used in this lesson, s(t) = t³ − 6t² + 9t, and compares the signs of velocity and acceleration to find when the particle is speeding up.Watch on YouTube

Displacement versus total distance

Displacement is the overall change in position, $s (end) - s (start)$ , so back-and-forth trips cancel. Total distance counts every metre travelled, whichever way. To get it, split the time interval at every instant where the object changes direction, find the distance on each piece as $∣ s (end of piece) - s (start of piece) ∣,$ and add. For $s (t) = t^{3} - 6 t^{2} + 9 t$ from $t = 0$ to $t = 4$ , the positions at $t = 0, 1, 3, 4$ are $0, 4, 0, 4$ : the object goes out $4$ m, back $4$ m, then out $4$ m again. The displacement is $4$ m, but the total distance is $12$ m.

Loading interactive…

Choose the Reversing preset (initial velocity 12 m/s, acceleration −4 m/s²) and press play. The object slows while v and a have opposite signs, stops where the velocity graph crosses zero at t = 3 s, then speeds up in the negative direction.

Vertical motion near Earth's surface

A height model like

h (t) = - 4.9 t^{2} + v_{0} t + h_{0}

(metres, seconds) has

a (t) = h^{''} (t) = - 9.8 m/s^{2}

, the acceleration due to gravity, pointing down. The maximum height happens where

h^{'} (t) = 0

, and the object hits the ground where

h (t) = 0

Motion: Displacement, Velocity and Acceleration

Key things to remember

At rest and changing direction

Speeding up or slowing down

Displacement versus total distance

Worked examples

Where marks get lost

Flashcards 8

Practice