Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: rectilinear motion (201-SN2-RE)

This chapter introduces no new differentiation technique: it applies the ones from the course to three functions of time. The difficulty lies elsewhere, in READING the signs, and that is where papers lose their marks, almost always in the same places.

This sheet is written for CEGEP students in the Science program (Sciences de la nature) taking Differential Calculus, 201-SN2-RE (formerly 201-NYA). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

The velocity tells you which DIRECTION the particle is going; the acceleration only tells you whether the velocity increases or decreases. Almost all lost marks come from reading “negative acceleration” as “slowing down”, and from giving a displacement where the question asked for a distance.

This chapter is part of 201-SN2-RE, Differential Calculus

The essentials

Three functions of the same time, each one speaking about the previous one

• $v(t)=s'(t)$ in metres per second, $a(t)=v'(t)=s''(t)$ in metres per second squared, the position being read in metres.
• $v$ speaks only about $s$ , $a$ speaks only about $v$ : the acceleration says NOTHING about the direction of motion, and the velocity says nothing about the position.
• The particle moves along a straight LINE. The curve of $s(t)$ is not its path; it is the graph of its position over time.
• Going back from $a$ to $s$ requires TWO initial conditions, a starting velocity and a starting position, because each antidifferentiation brings in a constant.

The curve gives the POSITION, not the path: the particle goes back and forth along a line. It moves forward until

t=1

, backward until

t=3

, then forward again, and the two turnarounds are the extrema of

s

Writing the three units next to the three functions on the first line of your paper takes five seconds and catches half of the conversion errors.

Direction of motion: the SIGN of $v$ , never its variation

• $v>0$ : the particle moves forward. $v<0$ : it moves backward. That is all, and it depends only on the sign.
• TURNAROUND at the times where $v$ is zero AND CHANGES SIGN. These times are exactly the local extrema of the position.
• A velocity that is zero without changing sign, such as $v(t)=(t-2)^{2}$ , produces no turnaround: the particle pauses and sets off again in the same direction.
• On the graph of $v$ , the particle moves forward where the curve is ABOVE the axis, not where it rises.

On the graph of

v(t)=3(t-1)(t-3)

, the particle moves forward where the curve is ABOVE the axis (green) and backward below it (orange). Turnarounds at

t=1

and

t=3

, where

v

changes sign.

Speeding up or slowing down: the PRODUCT $va$ decides

• The speed $|v|$ has derivative $\frac{va}{|v|}$ : it increases if and only if $v$ and $a$ have the SAME sign.
• $v>0$ and $a>0$ : moving forward and speeding up. $v<0$ and $a<0$ : moving backward and speeding up.
• $v>0$ and $a<0$ : moving forward and slowing down. $v<0$ and $a>0$ : moving backward and slowing down.
• In other words, a negative acceleration slows down a particle moving forward and speeds up a particle moving backward.

On a test, the question “when is the particle speeding up?” is handled with a sign chart of THREE rows: $v$ , $a$ , then the product.

Displacement, distance, and the free-fall formulas

• DISPLACEMENT over $[t_{1}, t_{2}]$ : simply $s(t_{2})-s(t_{1})$ , with its sign.
• TOTAL DISTANCE TRAVELLED: split at the turnaround times, take absolute values, add. The two agree only if $v$ keeps a constant sign.
• Free fall, axis pointing up: $y(t)=y_{0}+v_{0}t-\frac{g}{2}t^{2}$ , $v(t)=v_{0}-gt$ , and $a=-g$ CONSTANT, including at the top.
• Without the time: $v^{2}=v_{0}^{2}+2a\left(y-y_{0}\right)$ , valid ONLY at constant acceleration.
• Simple harmonic motion: $x(t)=A\cos(\omega t)$ gives $a=-\omega^{2}x$ , with period $T=\frac{2\pi}{\omega}$ ; the speed is greatest at equilibrium, the acceleration at the endpoints.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading “negative acceleration” as “the particle is slowing down”

the whole question, and often the next one

What not to write

“ $a(1.5)=-3<0$ , so the particle is slowing down at that time.”

What to write

“ $v(1.5)=-2.25<0$ and $a(1.5)=-3<0$ : same sign, so the speed INCREASES. The particle moves backward faster and faster.”

Between

t=1

and

t=2

, the velocity is negative AND so is the acceleration: the particle moves backward faster and faster. A negative acceleration therefore does not mean “slowing down”.

Why: The sign of $a$ says in which direction the velocity CHANGES, not whether it grows in absolute value. The product $va$ decides, so you must always look at both.

2. Giving the displacement when the question asks for the total distance travelled

the whole question

What not to write

“Distance travelled between $t=0$ and $t=4$ : $s(4)-s(0)=4-0=4$ metres.”

What to write

“The particle turns around at $t=1$ and $t=3$ : distance $=|4-0|+|0-4|+|4-0|=12$ metres, whereas the displacement is $4$ metres.”

s(t)=t^{3}-6t^{2}+9t

: the particle moves forward

4

, back

4

, forward

4

. Distance travelled

12

m; the displacement

s(4)-s(0)=4

m sees only the end point.

Why: The displacement is a difference, the distance a sum of absolute values. Without splitting at the turnarounds, the trips out and back cancel each other and the answer is always too small.

3. Believing that the acceleration is zero at the top of a free fall

2 marks, and faulty physical reasoning

What not to write

“At the highest point, the object stops, so $v=0$ and $a=0$ .”

What to write

“At the top $v=0$ , but $a=-g=-9.8$ metres per second squared, as everywhere else: it is precisely this acceleration that brings the object back down.”

At the top the velocity is zero, but the acceleration is still

-9.8

metres per second squared: it is what brings the object back down. A zero acceleration would leave it hanging in the air.

Why: In free fall the acceleration is CONSTANT throughout the motion. If it vanished at the top, the velocity would stay zero there and the projectile would float in the air.

4. Concluding there is a turnaround as soon as the velocity is zero

1 to 2 marks, and a distance travelled split for nothing

What not to write

“ $v(t)=(t-2)^{2}$ is zero at $t=2$ , so the particle turns around at that time.”

What to write

“ $v$ is zero at $t=2$ but stays positive on either side: the particle pauses and sets off again in the SAME direction. There is no turnaround.”

v(t)=(t-2)^{2}

touches the axis at

t=2

without crossing it: the velocity stays positive on both sides. The particle stops for an instant and sets off again in the same direction.

Why: A turnaround requires a CHANGE OF SIGN, not just a zero. The distinction is the same as between a critical number and an extremum, seen in the previous chapter.

5. Reading the velocity graph as if it were the position graph

the whole graph-reading question, that is 4 or 5 marks

What not to write

“The curve rises between $t=2$ and $t=3$ , so the particle moves forward on that interval.”

What to write

“This is the graph of $v$ : the particle moves forward where the curve is ABOVE the axis. Where it rises only means $a>0$ .”

Between

t=2

and

t=3

, the curve of

v

RISES (

a>0

) but stays below the axis (

v<0

): the particle moves backward, more and more slowly. The direction is read from the sign, not from the slope.

Why: On a graph of $v$ , the HEIGHT gives the direction of motion and the SLOPE gives the acceleration. Mixing them up means answering a question that was not asked.

6. Going back from the acceleration to the position with a single constant

2 marks, and a trajectory that starts from the wrong place

What not to write

“ $a(t)=6$ , so $v(t)=6t$ and $s(t)=3t^{2}$ .”

What to write

“ $v(t)=6t+v_{0}$ and $s(t)=3t^{2}+v_{0}t+s_{0}$ : two antidifferentiations, so TWO constants, to be determined by the two initial conditions given in the question.”

Why: Each antidifferentiation brings in a constant. Forgetting them amounts to assuming that the particle starts from the origin and at rest, which the question rarely says.

7. Using $v^{2}=v_{0}^{2}+2a\Delta y$ when the acceleration is not constant

the whole question

What not to write

“ $a(t)=6t-12$ , so $v^{2}=v_{0}^{2}+2a(s-s_{0})$ .”

What to write

“This formula assumes $a$ is CONSTANT. Here $a$ depends on time: go through $v(t)$ and $s(t)$ , or antidifferentiate.”

Why: The time-free formula is proved by integrating a constant acceleration. Used outside that setting, it gives a plausible and wrong number, which is the worst case.

8. Mixing kilometres per hour and metres per second

1 to 2 marks, and an absurd result off by a factor of

12.96

What not to write

“The car is travelling at $90$ kilometres per hour, so $v_{0}=90$ in the formula.”

What to write

“ $90$ kilometres per hour is $25$ metres per second, and all the formulas of the chapter are in metres per second.”

Why: The acceleration is in metres per second squared: mixing units throws the result off by a factor of $3.6$ to the first power, and of $3.6^{2}$ in a formula in $v^{2}$ .

Which method to choose

Which function to look at, depending on the verb of the question

Every question of the chapter is handled with $s$ , $v$ and $a$ ; the verb tells you which one to look at, and from which angle, sign or variation.

If “in which direction”, “when does the particle move forward or backward” → SIGN of $v$ , with a sign chart

Example: $v(t)=3(t-1)(t-3)$ is positive outside $[1, 3]$
If “when does it turn around”, “when does it change direction” → the zeros of $v$ where the sign CHANGES, to be checked on both sides

Example: $t=1$ and $t=3$ , but not $t=2$ for $v=(t-2)^{2}$
If “when is it speeding up”, “when is it slowing down” → sign of the PRODUCT $va$ , with a three-row chart

Example: $v<0$ and $a<0$ on $(1, 2)$ : the particle speeds up
If “displacement”, “how far has it moved from where it started” → $s(t_{2})-s(t_{1})$ , a simple signed difference

Example: $s(4)-s(0)=4$ metres
If “distance travelled”, “how far has it gone in all” → split at the turnarounds, absolute values, sum

Example: $|4|+|-4|+|4|=12$ metres

The distance is always greater than or equal to the absolute value of the displacement: that is the instant check.
If “the position, given the acceleration” → two antidifferentiations and two initial conditions

Example: $a=-9.8$ gives $y=y_{0}+v_{0}t-4.9t^{2}$

A question about SPEED, or about “how fast”, is about $|v|$ and not $v$ : the sign disappears, but the product $va$ remains the tool.

Which free-fall formula, depending on the data given

In free fall, four formulas and a single criterion: which quantity the question gives, and which one it does not.

If the question gives the time $t$ → $y(t)=y_{0}+v_{0}t-\frac{g}{2}t^{2}$ and $v(t)=v_{0}-gt$

Example: position and velocity at $t=3$ seconds
If the question neither gives the time nor asks for it → $v^{2}=v_{0}^{2}+2a(y-y_{0})$ , which skips the time

Example: landing speed given the starting height
If the question asks for the MAXIMUM HEIGHT → solve $v=0$ to find the time of the top, then substitute into $y(t)$

Example: $v_{0}=20$ gives $t=\frac{20}{9.8}\approx 2.04$ s and $y\approx 20.4$ m
If the question asks for the TOTAL TIME of flight → solve $y(t)=y_{\text{ground}}$ , keeping the positive root

Example: $20t-4.9t^{2}=0$ gives $t=0$ or $t\approx 4.08$ s

When the start and the landing are at the same height, the time going up equals the time coming down: that is the consistency check.

The acceleration is $-g$ during the WHOLE motion, on the way up, at the top and on the way down. None of the four formulas changes sign along the way.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Computing the distance travelled over an interval

When to use it: The question asks for the distance travelled, the total distance, or “how many metres the particle has covered”.

1 Compute $v(t)=s'(t)$ and FACTOR it: “ $v(t)=3t^{2}-12t+9=3(t-1)(t-3)$ .”
2 Find the turnaround times, checking the change of sign: “ $v$ is zero at $t=1$ and $t=3$ , and changes sign at both.”
3 Compute the position at each splitting time, endpoints included: “ $s(0)=0$ , $s(1)=4$ , $s(3)=0$ , $s(4)=4$ .”
4 Add the absolute values of the changes, piece by piece, showing the sum.
5 Compare with the displacement and check that the distance is greater than or equal to it.

Concluding sentence

The particle turns around at $t=1$ and $t=3$ ; the distance travelled over $[0, 4]$ is $|4-0|+|0-4|+|4-0|=12$ metres, for a displacement of only $4$ metres.

The trap: Leaving the endpoints $t=0$ and $t=4$ out of the list of computed positions. The first and last pieces then vanish, and the distance is too small with nothing to warn you.

Marking: Typically 1 mark for factoring $v$ , 1 mark for the turnaround times justified by the change of sign, 2 marks for the splitting and the sum.

Solving a free-fall problem

When to use it: An object thrown, dropped, launched upward: any motion with constant acceleration $-g$ .

1 Choose and STATE the orientation of the axis: “Take the axis pointing up, with the origin at the launch point.”
2 Write the three functions with their constants: “ $a=-9.8$ , $v(t)=20-9.8t$ , $y(t)=20t-4.9t^{2}$ .”
3 Translate the question into an equation: the top gives $v=0$ , landing gives $y=y_{\text{ground}}$ , a given velocity gives $v(t)=\ldots$
4 Solve, then substitute into the function asked for, never switching formulas along the way.
5 Check the order of magnitude and the sign: a negative height or a negative time signals a badly chosen root.

Concluding sentence

The projectile reaches its maximum height at $t=\frac{20}{9.8}\approx 2.04$ seconds, at a height of about $20.4$ metres, and lands after about $4.08$ seconds.

The trap: Changing the sign of $g$ on the way down, believing you are “putting the acceleration back in the right direction”. The axis is chosen once and for all at the start, and $a=-g$ during the whole motion.

Marking: 1 mark for the stated axis and the three functions, 1 mark per correctly translated equation, 1 mark for the answer with its unit.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Test the product $va$ at one point of each interval

Pick a value in the middle of each interval of the chart, compute $v$ and $a$ , and look at the SIGN of the product. Positive, the particle is speeding up; negative, it is slowing down.

At $t=1.5$ : $v=-2.25$ and $a=-3$ , positive product, so it is speeding up. At $t=2.5$ : $v=-2.25$ and $a=3$ , negative product, so it is slowing down.
The distance is always greater than or equal to the displacement

Compare the distance you found with the absolute value of the displacement. A smaller distance is necessarily wrong, and an equal distance means that $v$ did not change sign.

$12\geq|4|$ : consistent. A distance of $4$ metres when the particle turned around twice gives away a forgotten split.
Check the physical order of magnitude

Convert the result into everyday units before handing it in: a speed in kilometres per hour, a height in floors, a time in seconds you can feel.

A speed of $1100$ metres per second for a ball thrown by hand, nearly $4000$ kilometres per hour, signals a conversion or formula error.
Check the units by differentiating in your head

Differentiating divides by seconds: metres give metres per second, then metres per second squared. An answer whose unit does not follow this chain is wrong.

An acceleration given in metres per second signals that you gave $v$ instead of $a$ , a frequent mix-up at the end of a test.

The typical problem, taken apart

A complete motion over four seconds, from the signs to the distance

A particle moves along a straight line, its position in metres being $s(t)=t^{3}-6t^{2}+9t$ for $t$ in seconds, $t\in[0, 4]$ .

1. Find the turnaround times and the direction of motion.

2. Compute the displacement, then the distance travelled.

3. State on which intervals the particle is speeding up.

Step 1

$v(t)=s'(t)=3t^{2}-12t+9=3(t-1)(t-3)$ , so $v>0$ on $[0, 1)$ and $(3, 4]$ , and $v<0$ on $(1, 3)$ .

Why

FACTOR before drawing the sign chart: an expanded expression forces you to solve on each interval, a factored one is read at a glance.

Step 2

$v$ is zero at $t=1$ and $t=3$ , and changes sign at both: the particle therefore turns around at these two times.

Why

Being zero is not enough; the change of sign is what makes a turnaround. Saying so explicitly earns the rigour mark the marker is looking for here.

Step 3

$s(0)=0$ , $s(1)=1-6+9=4$ , $s(3)=27-54+27=0$ , $s(4)=64-96+36=4$ .

Why

Compute the position at the splitting times AND at both endpoints. Forgetting an endpoint makes a whole piece of the distance travelled disappear.

Step 4

Displacement: $s(4)-s(0)=4-0=4$ metres.

Why

The displacement is a simple difference, blind to the path followed. It is the answer too often given when the question asked for the distance.

Step 5

Distance: $|s(1)-s(0)|+|s(3)-s(1)|+|s(4)-s(3)|=|4|+|-4|+|4|=12$ metres.

Why

Three pieces because there are two turnarounds. The absolute value keeps the trip out and the trip back from cancelling each other, which is exactly what “travelled” means.

Step 6

$a(t)=6t-12$ , zero at $t=2$ . Chart of the product: $va<0$ on $[0, 1)$ and $(2, 3)$ , $va>0$ on $(1, 2)$ and $(3, 4]$ .

Why

Build the three-row chart, $v$ , $a$ , then the product. It is the only way not to go wrong on the two intervals where the velocity is negative.

Step 7

Check: $12\geq|4|$ , and at $t=1.5$ we have $v=-2.25$ and $a=-3$ , of the same sign, which confirms that the particle speeds up on $(1, 2)$ .

Why

Two independent checks in ten seconds: the distance versus displacement inequality, and a test point in the most counterintuitive interval of question 3.

The conclusion, written out

The particle turns around at $t=1$ and $t=3$ ; its displacement over $[0, 4]$ is $4$ metres and the distance travelled $12$ metres; it speeds up on $(1, 2)$ and on $(3, 4]$ , and slows down on $[0, 1)$ and $(2, 3)$ .

The classic mistake on this problem: Answering that the particle slows down on $(1, 2)$ because the acceleration is negative there. The velocity is negative there too, so the product is positive and the particle moves backward faster and faster: this is precisely the interval the test is aiming at.

Learn by heart

• $v=s'$ , $a=v'=s''$ : each derivative speaks only about the function just before it.
• The SIGN of $v$ gives the direction; the SLOPE of $v$ gives the acceleration. On a graph of $v$ , never confuse the two.
• Turnaround only if $v$ is zero AND CHANGES SIGN.
• The particle speeds up if and only if $v$ and $a$ have the SAME sign: the product $va$ decides.
• Displacement $=s(t_{2})-s(t_{1})$ ; distance $=$ sum of the absolute values after splitting at the turnarounds.
• The distance is always greater than or equal to the absolute value of the displacement.
• In free fall, $a=-g$ EVERYWHERE, top included, and $v^{2}=v_{0}^{2}+2a\Delta y$ holds only at constant acceleration.
• Going back from $a$ to $s$ requires TWO constants, so two initial conditions.

Frequently asked questions

How do you know if a particle is speeding up or slowing down?

Look at the product of the velocity and the acceleration. If they have the same sign, the speed increases and the particle is speeding up; if they have opposite signs, it is slowing down. A negative acceleration alone is therefore not enough to conclude: it slows down a particle moving forward, but it speeds up a particle moving backward.

What is the difference between displacement and distance travelled?

The displacement is the difference between the final position and the starting position, with its sign: it ignores the path followed. The distance travelled adds up the lengths of all the pieces, in absolute value, after splitting the trip at the turnaround times. The two agree only if the particle never changes direction.

When does a particle turn around?

At the times when the velocity is zero and changes sign. Being zero is not enough: a velocity that touches zero without changing sign, like a perfect square, corresponds to a pause followed by a departure in the same direction. These turnaround times are exactly the local extrema of the position.

Is the acceleration zero at the top of a trajectory?

No. At the top, it is the velocity that is zero, not the acceleration. In free fall the acceleration is minus nine point eight metres per second squared during the whole motion, on the way up, at the top and on the way down. If it vanished at the top, the object would hang motionless in the air instead of falling back.

How do you find the position when you know the acceleration?

By antidifferentiating twice, which brings in two constants. The first antidifferentiation gives the velocity up to a constant, fixed by the initial velocity; the second gives the position up to a constant, fixed by the initial position. A question that supplies only one initial condition is incomplete.

Practise it

Corrected exercises: Rectilinear motion, 201-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

16 corrected exercises
160 points
225 minutes

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-rectilinear-motion. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: rectilinear motion (201-SN2-RE)

The essentials

Three functions of the same time, each one speaking about the previous one

Direction of motion: the SIGN of vvv, never its variation

Speeding up or slowing down: the PRODUCT vavava decides

Displacement, distance, and the free-fall formulas

The mistakes that cost marks

1. Reading “negative acceleration” as “the particle is slowing down”

2. Giving the displacement when the question asks for the total distance travelled

3. Believing that the acceleration is zero at the top of a free fall

4. Concluding there is a turnaround as soon as the velocity is zero

5. Reading the velocity graph as if it were the position graph

6. Going back from the acceleration to the position with a single constant

7. Using v2=v02+2aΔyv^{2}=v_{0}^{2}+2a\Delta yv2=v02​+2aΔy when the acceleration is not constant

8. Mixing kilometres per hour and metres per second

Which method to choose

Which function to look at, depending on the verb of the question

Which free-fall formula, depending on the data given

How the answer is expected to be written

Computing the distance travelled over an interval

Solving a free-fall problem

Check before you hand in

The typical problem, taken apart

A complete motion over four seconds, from the signs to the distance

Learn by heart

Frequently asked questions

How do you know if a particle is speeding up or slowing down?

What is the difference between displacement and distance travelled?

When does a particle turn around?

Is the acceleration zero at the top of a trajectory?

How do you find the position when you know the acceleration?

Corrected exercises: Rectilinear motion, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?

Direction of motion: the SIGN of $v$ , never its variation

Speeding up or slowing down: the PRODUCT $va$ decides

7. Using $v^{2}=v_{0}^{2}+2a\Delta y$ when the acceleration is not constant