Lesson 3 of 5Kinematics

Unit 1 · Lesson 3 of 5

Acceleration and the Kinematic Equations

What acceleration really measures, why a negative one does not always mean slowing down, and how to pick the right equation out of the four on your formula sheet.

4 min read10 practice questionsInteractiveAsk AI about this →

Key things to remember

1 $a = \frac{Δ v}{Δ t} = \frac{v _{f} - v _{i}}{Δ t}$ , measured in $m/s^{2}$ .
2Same signs on velocity and acceleration means speeding up; opposite signs means slowing down.
3 $v_{f} = v_{i} + a Δ t$ (no displacement) and $v_{f}^{2} = v_{i}^{2} + 2 a Δ d$ (no time).
4 $Δ d = v_{i} Δ t + \frac{1}{2} a Δ t^{2}$ (no final velocity) and $Δ d = v_{f} Δ t - \frac{1}{2} a Δ t^{2}$ (no initial velocity).
5Memorize the off-sheet one: $Δ d = (\frac{v _{i} + v _{f}}{2}) Δ t$ , which is missing acceleration.
6These four equations are only valid while the acceleration stays constant. Split the problem at any point where it changes.

Acceleration is the rate at which velocity changes. That one sentence contains the whole idea, but it hides two traps. Velocity is a vector, so it changes if the speed changes, if the direction changes, or both. And because acceleration is itself a vector, its sign tells you which way the velocity is being pushed — not whether the object is getting faster or slower.

a = \frac{Δ v}{Δ t} = \frac{v _{f} - v _{i}}{Δ t}

Change in velocity per unit time. The unit is metres per second, per second.

An acceleration of $3.0 m/s^{2}$ means the velocity gains $3.0 m/s$ every second. Start at rest and after one second you are doing $3.0 m/s$ , after two seconds $6.0 m/s$ , and so on. That reading of the unit — 'metres per second, every second' — is worth saying out loud until it is automatic.

Negative acceleration does not mean slowing down

Compare the signs of velocity and acceleration, not the sign of acceleration alone. If they have the same sign the object is speeding up; if they have opposite signs it is slowing down. A car reversing and picking up speed has a negative velocity and a negative acceleration, and it is definitely getting faster. The word for slowing down is 'decelerating', and it is a description, not a sign.

Velocity	Acceleration	What is happening
$+$	$+$	Moving forward, speeding up
$+$	$-$	Moving forward, slowing down
$-$	$-$	Moving backward, speeding up
$-$	$+$	Moving backward, slowing down
Any	$0$	Constant velocity — uniform motion

Uniformly accelerated motion

Everything below assumes uniform acceleration — an acceleration that stays constant over the interval. That is a real restriction, and it is why these equations cannot be used across a stretch where the acceleration changes: split such a problem into intervals and treat each one separately.

Fig 1Five variables, four equations. Each equation is missing exactly one variable — find the one you neither have nor want.

There are five quantities in play: $v_{i}$ , $v_{f}$ , $a$ , $Δ t$ and $Δ d$ . Any problem gives you three and asks for a fourth, which leaves one quantity that is neither given nor wanted. Each equation on the formula sheet is missing exactly one of the five, so the equation you want is the one missing the variable you do not care about. This turns equation choice from guesswork into a lookup.

v_{f} = v_{i} + a Δ t

Missing displacement.

Δ d = v_{i} Δ t + \frac{1}{2} a Δ t^{2}

Missing final velocity.

Δ d = v_{f} Δ t - \frac{1}{2} a Δ t^{2}

Missing initial velocity.

v_{f}^{2} = v_{i}^{2} + 2 a Δ d

Missing time — the one to reach for when no time is given or wanted.

The fifth equation is not on your sheet

Δ d = (\frac{v _{i} + v _{f}}{2}) Δ t

— displacement as average velocity times time — is genuinely useful and is missing from the formula sheet, so memorize it. It is the one that is missing acceleration, which makes it the quickest route whenever you know both velocities and the time. It only works for uniform acceleration, because only then is the average velocity the midpoint of the two.

A method that does not fail you under pressure

Draw the motion and mark a positive direction. Everything after this depends on it.
List all five variables down the side of the page, filling in what you know, writing a question mark beside what you want, and leaving one blank.
Choose the equation missing the blank one. No rearranging yet.
Rearrange algebraically first, substitute second. Substituting first and then trying to untangle the numbers is where arithmetic errors breed.
Check the sign and the size. Does the direction make sense? Is $47 m/s$ a plausible speed for a dropped ball?

When time is the unknown in the squared equation

Solving

Δ d = v_{i} Δ t + \frac{1}{2} a Δ t^{2}

for time gives a quadratic. Put it in the form

\frac{1}{2} a Δ t^{2} + v_{i} Δ t - Δ d = 0

and use the quadratic formula from MCR3U. You will usually get two roots; discard the negative one, since negative time has no meaning here.

Loading interactive…

Change the acceleration and watch which graph responds: the velocity-time slope changes, and the position-time line bends.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A car accelerates uniformly from rest to

25 m/s

8.0 s

. Find the acceleration and the distance covered.

Example 2

A cyclist moving at

12 m/s

brakes uniformly and stops in

18 m

. Find the acceleration.

Example 3

A train starts at

8.0 m/s

and accelerates at

1.5 m/s^{2}

for

12 s

. How far does it travel, and how fast is it going at the end?

Where marks get lost

Reading a negative acceleration as 'slowing down'. Compare it with the sign of the velocity — same signs speed up, opposite signs slow down.
Squaring only the number and not the unit, so $\frac{1}{2} a Δ t^{2}$ comes out with the wrong power of seconds and the error goes unnoticed.
Using a kinematic equation across an interval where the acceleration changed. They are valid only for uniform acceleration — split the journey up.
Substituting numbers before rearranging, then trying to unpick a squared term buried inside an expression.
Forgetting that $v_{f} = 0$ for anything that comes to rest, and then claiming there is not enough information.
Taking the positive root of a quadratic for $Δ t$ without checking, or keeping a negative time as a valid answer.

Flashcards 9

Acceleration

How do you know an object is speeding up?

Equation missing displacement

Equation missing time

+ 5 more

Practice

A car speeds up from

10.0 m/s

30.0 m/s

5.00 s

. What is its acceleration in

m/s^{2}

m/s^2

An object has a velocity of

- 12 m/s

and an acceleration of

- 3.0 m/s^{2}

. What is it doing?

A sled starts from rest and accelerates at

2.0 m/s^{2}

for

6.0 s

. How far does it travel, in metres?

A puck slows uniformly from

18 m/s

6.0 m/s

over

24 m

. What is the magnitude of its acceleration in

m/s^{2}

m/s^2

You know

v_{i}

v_{f}

and

Δ t

, and you want

Δ d

. Which equation is the most direct route?

What condition must hold for the four kinematic equations to be valid?

A motorcycle accelerates from rest at

4.5 m/s^{2}

. What is its speed, in m/s, after covering

50.0 m

m/s

Match each kinematic equation to the variable it does not contain.

v_{f} = v_{i} + a Δ t

v_{f}^{2} = v_{i}^{2} + 2 a Δ d

Δ d = v_{i} Δ t + \frac{1}{2} a Δ t^{2}

Δ d = v_{f} Δ t - \frac{1}{2} a Δ t^{2}

Which of these is not printed on the SPH3U formula sheet?

Order the steps for solving a kinematics problem reliably.

1Check the sign and whether the size is physically plausible
2Rearrange algebraically, then substitute the numbers
3Pick the equation that is missing the blank variable
4Draw the motion and choose a positive direction
5List all five variables, marking the known, the wanted and the blank

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