PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: motion in a straight line (PHYS 101)

This revision sheet is not a summary of the chapter, you already have the course notes. It answers one question: what makes a student lose marks on straight line motion in PHYS 101, the calculus-free mechanics course at McGill University, and what exact gesture prevents each loss.

Everything here is done with algebra and geometry. An instantaneous velocity is the slope of a tangent read on a graph, an area under a velocity-time curve is cut into triangles and rectangles, and no derivative or integral appears anywhere.

Updated on October 25, 2026

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The thread of the chapter

The sign is not decoration and it is not the direction of travel: it is the answer to a question you must ask before the first calculation, which way does my axis point. Once the axis is written down, the sign of the velocity gives the direction of motion and the sign of the acceleration gives the direction in which that velocity is being changed, nothing more.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Displacement and distance travelled
Average velocity and average speed
Instantaneous velocity
Sign convention on an axis
Average and instantaneous acceleration
Equations of uniformly accelerated motion
Free fall acceleration
Area under a velocity-time graph
Slope of a position-time graph
Meeting point of two mobiles
Motion in two phases
Dimensional analysis and limiting cases

The essentials

On your formula sheet: this chapter's lines, and when each one holds

• $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ : Write the kinematic equation as $\frac{1}{2}at^{2}+v_{0}t+(x_{0}-x)=0$ first: the $a$ of the formula is then $\frac{1}{2}a$ of the motion. Keep the root that makes physical sense and say in one phrase why the other is rejected.
• $g=9.81$ : In m/s², the magnitude of the free-fall acceleration near the Earth's surface. Its sign comes from your axis: $a=-g$ when up is positive.
• $v=v_{0}+at$ : Constant acceleration only. No displacement in it: the line to use when $\Delta x$ is neither given nor asked.
• $x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$ : Constant acceleration only, and no final velocity in it. Every letter carries the sign of your axis.
• $v^{2}=v_{0}^{2}+2a(x-x_{0})$ : Constant acceleration only, and no time in it. It gives $v^{2}$ : the sign of $v$ comes from the situation, not from the square root.

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam. This sheet writes $v_{i}$ and $v_{f}$ where the formula sheet writes $v_{0}$ and $v$ : same quantities.

The four equations, sorted by what is MISSING

• Without the displacement: $v_{f} = v_{i} + a\,\Delta t$ .
• Without the final velocity: $\Delta x = v_{i}\,\Delta t + \tfrac{1}{2}a\,\Delta t^{2}$ .
• Without the time: $v_{f}^{2} = v_{i}^{2} + 2a\,\Delta x$ .
• Without the acceleration: $\Delta x = \dfrac{v_{i} + v_{f}}{2}\,\Delta t$ .
• All four require a CONSTANT acceleration, that is, a straight line on the velocity-time graph.

Do not pick an equation by the theme of the problem. List the five quantities $v_{i}$ , $v_{f}$ , $a$ , $\Delta t$ , $\Delta x$ , cross out the three the statement gives you and the one it asks for: the quantity left over names the equation. This reflex is worth one or two lines on every problem of the chapter, and it removes the intermediate value that most sign errors are built on.

The sign of a tells you nothing about speeding up

• The sign of $v$ answers: which way is the object going along MY axis.
• The sign of $a$ answers: which way is the velocity being changed along MY axis.
• Same signs, the speed grows. Opposite signs, the speed falls. There is no third case.
• A speed is never negative. A velocity may be, and reporting one as the other loses the mark even when the number is right.

The same acceleration

a = +3

m/s² appears in case 1, where the object speeds up, and in case 4, where it slows down: only the comparison with the sign of

v

decides, never the sign of

a

alone.

Write the convention in words on the first line, I take the upward direction as positive, therefore $a = -9.8$ m/s². One line, and the marker can then tell a deliberate minus sign from a slip. Without it, every sign in the rest of the page is unverifiable.

Free fall: one acceleration for the whole flight

• Once the ball has left the hand, $a = -9.8$ m/s² (upward positive) on the way up, at the top and on the way down.
• At the top $v = 0$ for an instant while $a$ is still $-9.8$ m/s²: a zero velocity is not a zero acceleration.
• Thrown up at $v_{0}$ and caught at the same height, the ball climbs for $\frac{v_{0}}{g}$ , falls for the same time, and comes back at $-v_{0}$ .

A ball thrown up at

19.6

m/s, every

0.5

s, spread sideways: the teal velocity shrinks by equal steps, passes through zero at the top and grows downward, while the brown acceleration never changes.

Air resistance is left out everywhere in PHYS 101 unless a statement brings it in.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which equation, according to the quantity that is missing

Read the line whose first cell names the quantity that appears neither in the data nor in the question. The last column is the equation to write down, and the example underneath is the one-line calculation it produces.

What is ABSENT	What you must already know	The equation
the displacement $\Delta x$	$v_{i}$ , $a$ and $\Delta t$	$v_{f} = v_{i} + a\,\Delta t$
Example: A cart launched up a tilted air track at $6.0$ m/s, with $a = -1.5$ m/s², after $8.0$ s: $6.0 - 12.0 = -6.0$ m/s, it is gliding back down.
the final velocity $v_{f}$	$v_{i}$ , $a$ and $\Delta t$	$\Delta x = v_{i}\Delta t + \tfrac{1}{2}a\Delta t^{2}$
Example: Same cart over the same $8.0$ s: $48 - 48 = 0$ m of displacement, and yet $24$ m travelled.
the time $\Delta t$	$v_{i}$ , $v_{f}$ and $a$	$v_{f}^{2} = v_{i}^{2} + 2a\,\Delta x$
Example: Braking from $25$ m/s at $-6.5$ m/s²: $\Delta x = \dfrac{625}{13} = 48$ m.
the acceleration $a$	$v_{i}$ , $v_{f}$ and $\Delta t$	$\Delta x = \dfrac{v_{i} + v_{f}}{2}\Delta t$
Example: Stopping from $25$ m/s in $4.0$ s: $\dfrac{25 + 0}{2}(4.0) = 50$ m, in one line.
a value hidden in the WORDS	from rest, the top, stops or turns around: each one means $v = 0$	write $v = 0$ down, then use the row of what is still missing
Example: The cart turns around when $v_{f} = 0$ : $0 = 6.0 - 1.5\,\Delta t$ gives $\Delta t = 4.0$ s, then $\Delta x = \dfrac{6.0 + 0}{2}(4.0) = 12$ m up the track.

The fourth line is the one students forget, and it is the cheapest of the four: no acceleration to compute, no intermediate value to carry. The last line is the datum the statement gives in words rather than in numbers, and PHYS 101 problems expect it on every line: from rest, highest point, comes to a stop all write $v = 0$ . The four equations use $\Delta x$ , a DISPLACEMENT, never a distance travelled.

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 a negative acceleration as a reverse gear

2 marks, and the rest of the problem is then solved in the wrong direction

What not to write

“ $a = -1.5$ m/s², so the cart is going back down the track.”

What to write

“ $a < 0$ and $v > 0$ : the cart is still going up the track and slowing down. It only moves back down after $t = 4.0$ s, when $v$ itself becomes negative.”

Why: The sign of $a$ says which way the velocity is being changed, not which way the object is going. The two signs agree whenever the object is speeding up and disagree whenever it is slowing down, so neither can be read without the other.

2. Reporting the area under a velocity-time graph as the distance travelled

the whole question, 3 or 4 marks, because both numbers are then wrong in the next part

What not to write

“The area is $16 - 4 = 12$ , so the cart travelled $12$ m.”

What to write

“The signed area gives the displacement, $+12$ m. The distance travelled adds the areas without their signs, $16 + 4 = 20$ m.”

The two triangles have the same area and opposite signs: the displacement over the eight seconds is

0

m and the distance travelled is

40

m, for one and the same motion.

Why: A region below the time axis is a piece of motion in the negative direction. It subtracts from the displacement and adds to the distance, which is exactly why the two answers differ as soon as the graph crosses the axis.

3. Averaging two velocities instead of dividing the displacement by the time

3 marks, and the value obtained is usually not even close

What not to write

“She walks $30$ m out and $20$ m back, so her average velocity is the average of $+1.5$ and $-1.0$ m/s.”

What to write

“ $v_{av} = \dfrac{\Delta x}{\Delta t} = \dfrac{+10}{50} = +0.20$ m/s: total displacement over total elapsed time, including the $10$ s spent standing still.”

Why: Averaging two velocities weights them equally, which is legitimate only when they last equally long. The definition never is wrong: displacement divided by elapsed time.

4. Using $g = -9.8$ m/s² with an axis that points downwards

1 mark at once, and often the whole question, since a negative depth is rarely noticed

What not to write

“Depths are measured positively downwards, and $a = -9.8$ m/s², so $\Delta x = -4.9\Delta t^{2}$ .”

What to write

“Downwards is positive, therefore $a = +9.8$ m/s² and $\Delta x = +4.9\Delta t^{2}$ , a positive depth.”

Why: The number $9.8$ is a magnitude and it never changes. Its sign is a consequence of the axis you chose, so it cannot be memorised, it has to be decided once per problem and written down.

5. Concluding that the acceleration is zero at the top of a throw

2 marks, and it is asked almost every term

What not to write

“At the highest point $v = 0$ , so $a = 0$ there.”

What to write

“At the highest point $v = 0$ but $a = -9.8$ m/s², unchanged: the velocity is zero for one instant and it is still being changed, which is precisely why the ball comes back down.”

Why: The acceleration measures how fast the velocity is CHANGING, not how large it is. A velocity of zero at an instant says nothing at all about the acceleration at that instant.

6. Using the midpoint formula when the acceleration is not constant

the whole question, and the absurd answer is usually handed in as it stands

What not to write

“The cart starts at rest and ends at rest, so $v_{av} = \dfrac{0 + 0}{2} = 0$ .”

What to write

“ $v_{av} = \dfrac{\Delta x}{\Delta t} = \dfrac{39}{9.0} = 4.3$ m/s. The formula $\dfrac{v_{i}+v_{f}}{2}$ holds over one phase of constant acceleration, not over three.”

Why: That formula is a consequence of the velocity growing linearly, not a definition. As soon as the velocity-time graph is bent or broken, only the definition survives.

7. Handing in a negative velocity as a speed

half a mark to one mark each time, and it recurs on every part of the question

What not to write

“The ball comes back past the hand at a speed of $-19.6$ m/s.”

What to write

“Its velocity is $-19.6$ m/s, so its speed is $19.6$ m/s, the same as at launch, in the opposite direction.”

Why: Speed is the magnitude of the velocity, so it is never negative. Keeping both words apart on the copy also keeps them apart in the reasoning, which is where the real saving is.

8. Dividing one position by one time to get an instantaneous velocity

2 marks, and the wrong value is plausible enough to survive every later check

What not to write

“At $t = 3.0$ s the glider is at $x = 6.5$ m, so $v = \dfrac{6.5}{3.0} = 2.2$ m/s.”

What to write

“ $v$ is the slope of the TANGENT at that point: read two points ON the tangent, $(1.0 ; 0.5)$ and $(5.0 ; 12.5)$ , so $v = \dfrac{12.0}{4.0} = 3.0$ m/s.”

Why: A velocity is a ratio of two CHANGES, so it needs two readings. Dividing $x$ by $t$ is only legitimate when the motion began at the origin at $t = 0$ and never changed speed.

9. Letting friction turn a motion around

the whole part, and every later answer built on the

-6.0

m/s

What not to write

“The puck slides at $6.0$ m/s and the ice slows it at $1.5$ m/s², so after $8.0$ s it is back at its start, moving at $-6.0$ m/s.”

What to write

“Friction can only stop the puck: it stops at $t = 4.0$ s after $12$ m and stays there. The equations hold only until $v = 0$ . A cart on a tilted air track does come back, because the slope still pulls once it has stopped.”

Same start and same

-1.5

m/s² until

t = 4.0

s: the cart on the tilted air track comes back to

x = 0

t = 8.0

s, the puck on ice stops at

12

m and stays.

Why: The four equations assume ONE acceleration over the whole interval. Before carrying them past a stop, ask whether the cause of the acceleration survives the stop: gravity along a slope does, friction does not.

10. Taking the instant the motors fail as the top of the climb

the whole part: a highest altitude

3.3

m too low, and every instant after the fault wrong

What not to write

“The motors fail at $16$ m, so $16$ m is the highest point and the drone starts down from rest.”

What to write

“At the fault the drone still rises at $8.0$ m/s: it climbs $0.82$ s more, to $19.3$ m, and only passes $16$ m again $0.82$ s after that, moving down at $8.0$ m/s.”

The drone's altitude, blue while the motors push at

2.0

m/s², orange once they fail: the curve keeps rising past

16

m, peaks at

19.3

m with a horizontal tangent, and is back at

16

m at

t = 5.63

Why: A motor that stops changes the ACCELERATION at once, never the velocity: the velocity at the end of one phase is the initial velocity of the next, and the top is where $v = 0$ , not where the motor stops.

Which method to choose

Which equation of motion to write down

List the five quantities $v_{i}$ , $v_{f}$ , $a$ , $\Delta t$ and $\Delta x$ . The statement gives three and asks for one. The fifth, the one that appears nowhere, names the equation.

If the TIME appears nowhere → $v_{f}^{2} = v_{i}^{2} + 2a\,\Delta x$

Example: braking from $25$ m/s at $-6.5$ m/s² gives $\Delta x = 48$ m

the most profitable equation of the chapter: braking distances and free fall heights almost never mention a duration
If the DISPLACEMENT appears nowhere → $v_{f} = v_{i} + a\,\Delta t$

Example: at $a = -9.8$ m/s² a ball thrown at $19.6$ m/s reaches the top after $2.0$ s
If the FINAL VELOCITY appears nowhere → $\Delta x = v_{i}\Delta t + \tfrac{1}{2}a\Delta t^{2}$

Example: a ruler released from rest falls $0.18$ m in $0.19$ s

with $v_{i} = 0$ it collapses to $\Delta x = \tfrac{1}{2}a\Delta t^{2}$ , the free fall workhorse
If the ACCELERATION appears nowhere → $\Delta x = \dfrac{v_{i} + v_{f}}{2}\,\Delta t$

Example: stopping from $25$ m/s in $4.0$ s covers $50$ m
If the statement describes TWO objects → write $x_{1}(t)$ and $x_{2}(t)$ , then set them equal

Example: $0.60t^{2} = 6.0t$ gives $t = 10$ s and $x = 60$ m

no single equation can do this: one origin, one clock, two position expressions

Two habits save the marks the arbre cannot. Convert every unit before the first line, and reject a root by naming the reason, a negative time before the motion started is outside the problem, not merely ugly.

What to do with a graph: slope or area

Look at what the vertical axis carries, then at what the question asks. You go down from position to velocity to acceleration by taking SLOPES, and back up by taking AREAS. There is never a third possibility.

One motion, three graphs: each is the SLOPE of the one before, and the shaded AREA under

v(t)

9 + 18 = 27

m, is the climb of

x(t)

If position-time graph, the velocity is asked → slope of the TANGENT at that instant

Example: a tangent through $(1.0 ; 0.5)$ and $(5.0 ; 12.5)$ gives $3.0$ m/s
If position-time graph, the AVERAGE velocity is asked → slope of the chord between the two points

Example: from $2.5$ m to $14.5$ m in $4.0$ s gives $3.0$ m/s

with a constant acceleration the chord equals the tangent at the MIDPOINT of the interval, which replaces a tangent nobody drew for you
If velocity-time graph, the acceleration is asked → slope of the segment

Example: from $+8.0$ to $-4.0$ m/s in $6.0$ s gives $-2.0$ m/s²
If velocity-time graph, the displacement is asked → signed AREA, cut into triangles and rectangles

Example: $+16$ m above the axis and $-4$ m below it give $+12$ m
If acceleration-time graph, the velocity is asked → AREA, added to the velocity you already had

Example: $2.0$ m/s² during $3.0$ s adds $6.0$ m/s to the initial value

an area on this graph is a CHANGE of velocity, never the velocity itself: the initial value has to come from the statement

Areas are cut into triangles and rectangles, always: this course has no integral, and none is needed as long as the graph is made of straight segments.

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.

Solving a problem of uniformly accelerated motion

When to use it: The statement gives three of the five kinematic quantities and asks for a fourth, with a constant acceleration.

1 Write the axis and the origin in words: I take the direction of the initial motion as positive, with the origin at the starting point.
2 List the data with their SIGNS and their units converted to metres and seconds, $g$ included.
3 Name the missing quantity and write the corresponding equation in letters, before any number.
4 Isolate the unknown in letters, then substitute, then compute once.
5 Write the answer as a sentence, with its unit and, if it is a velocity, with its sign and its direction.

Concluding sentence

“Taking the direction of travel as positive, the car stops after $\Delta x = 48$ m of braking.”

The trap: The step that gets dropped is always the first. Without a written convention the marker cannot tell whether your $-9.8$ is a choice or a slip, and the benefit of the doubt goes the other way.

Marking: 1 mark for the convention and the signed data, 1 mark for the equation in letters, 2 marks for isolating and computing, 1 mark for the concluding sentence with its unit.

Two objects that meet

When to use it: The statement says catch up, overtake, meet or cross, or gives one of the two a head start in time or in position.

1 Choose ONE origin of positions and ONE origin of time, shared by both objects, and say so.
2 Write $x_{1}(t)$ and $x_{2}(t)$ in full, putting any head start INSIDE the expression, for instance $t - 2.0$ for an object that leaves two seconds later.
3 Set $x_{1}(t) = x_{2}(t)$ and solve, then substitute the root back into BOTH expressions as a check.
4 Discuss every root: keep the ones that lie inside the situation, reject the others by naming the reason.

Concluding sentence

“The cyclist catches the runner $10$ s after the start, $60$ m from the line, a position confirmed by both expressions.”

The trap: Subtracting the head start at the end of the calculation instead of putting it inside the expression. The term in $\Delta t^{2}$ makes that correction simply false.

Marking: 2 marks for the two complete position expressions, 2 marks for solving, 1 mark for the reasoned rejection of the root that does not belong.

A motion in two phases

When to use it: An engine, a motor or a brake acts for a while and then stops, so the acceleration takes two values, one per phase.

1 Sketch the velocity-time graph first: one straight segment per phase, joined without a jump at the switch.
2 Phase 1 with its own acceleration: compute the velocity and the position at the switch.
3 Phase 2 starts from THOSE two values with the new acceleration. Never carry one equation across the switch.
4 Locate the top where $v = 0$ , then any later event by the displacement the statement gives for phase 2.
5 Read the answer back on the graph: after the switch, a triangle above the axis is a climb, a triangle below it a descent, and two that cancel bring the object back to the altitude of the switch.

Concluding sentence

“Taking upward as positive, the drone keeps climbing $0.82$ s after the fault, to $19.3$ m, and is back at the altitude of the fault at $t = 5.63$ s, moving down at $8.0$ m/s.”

The trap: Starting phase 2 from rest, or taking the altitude of the switch as the top. The junction hands the velocity over unchanged; only the slope of $v(t)$ breaks.

Marking: 1 mark for the graph and the convention, 2 marks for phase 1 and the values handed over, 2 marks for phase 2 and its conclusion, with the reading on the graph as the check of the last line.

Check before you hand in

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

Does the sign tell the same story as the words

If $a$ and $v_{i}$ have opposite signs, the speed must have DROPPED. If your final velocity is larger in magnitude than the initial one, a sign is wrong somewhere, and it is usually in the substitution rather than in the equation.

$v_{i} = +25$ m/s with $a = -6.5$ m/s² giving $v_{f} = +30$ m/s is impossible; the same data giving $v_{f} = -30$ m/s after a very short time is impossible too.
The order of magnitude, in one second

Keep three benchmarks in mind: a car brakes over a few tens of metres, a body released from rest falls about $5$ m in the first second, and a human reaction time is about two tenths of a second.

A stopping distance of $480$ m says kilometres per hour were left in the formula; a fall of $50$ m in one second says $g$ was used twice.
Displacement or distance, which word was in the question

Underline the word before answering. If the motion changes direction at any point, the two numbers differ, and the one you computed by adding signed areas is the displacement.

The cart that glides back down the air track to its launch point has $\Delta x = 0$ m and $24$ m travelled: answering $0$ m to a question about distance loses the whole part.
Is the acceleration really constant over the interval used

Before using any of the four equations, check that the velocity-time graph is a single straight segment over the interval. A broken line means one calculation per phase, joined by the velocity at each junction.

Over three phases lasting $9.0$ s in total, the midpoint formula gives $0$ m/s while the true average velocity is $4.3$ m/s.
A letter answer: its unit, a limit case, its direction of change

Before substituting numbers, check that the expression has the unit of the quantity asked, that a limit case gives back an answer you already know, and that it moves the right way when one data value grows.

$t = \dfrac{v_{0} + \sqrt{v_{0}^{2} + 2gh}}{g}$ is in seconds, tends to $\dfrac{2v_{0}}{g}$ as $h \to 0$ , and grows with $h$ . With $-2gh$ under the root it would shrink as the ball falls further: the sign is wrong.
Does the position graph tell the same story as the motion

On a position-time graph the slope is the velocity: a horizontal tangent where the object turns around, a straight piece where the acceleration is zero, a curve bending downwards where it is negative.

The cart launched up a tilted air track draws one downward parabola, flat at $t = 4.0$ s, back to $x = 0$ at $8.0$ s. Two straight lines meeting in a corner, or a curve that levels off and stays, describe another motion.

The typical problem, taken apart

A hospital lift, in three phases

A lift starts from rest on the ground floor and rises with a constant acceleration of $1.2$ m/s² for $2.0$ s.

It then travels at constant velocity for $4.0$ s, and finally slows uniformly to rest in $3.0$ s.

Find its maximum velocity, the acceleration of the last phase, the total height climbed and the average velocity of the trip.

The trip is one trapezium: the three areas

2.4

9.6

m and

3.6

m add up to the

15.6

m climbed, and the lift is going UP during all three, including the last one where the acceleration is negative.

Step 1

Upward is positive, origin at the ground floor. Phase 1: $v = v_{i} + a\,\Delta t = 0 + (1.2)(2.0) = 2.4$ m/s.

Why

The convention comes before any number, and the equation chosen is the one without $\Delta x$ , which the statement neither gives nor asks for at this stage.

Step 2

Phase 3: $a_{3} = \dfrac{0 - 2.4}{3.0} = -0.80$ m/s².

Why

The minus sign is not a descent. The lift is still going up, its velocity is still positive, and only the SPEED is falling. Answering $+0.80$ m/s² here costs the mark for the sign even though the magnitude is right.

Step 3

Displacements by areas: $\tfrac{1}{2}(2.0)(2.4) = 2.4$ m, then $(4.0)(2.4) = 9.6$ m, then $\tfrac{1}{2}(3.0)(2.4) = 3.6$ m.

Why

Three phases, three constant accelerations, so three separate calculations. Cutting the trapezium into a triangle, a rectangle and a triangle replaces every equation here and cannot go wrong on a sign, since all three regions lie above the axis.

Step 4

Total: $\Delta x = 2.4 + 9.6 + 3.6 = 15.6$ m over $\Delta t = 9.0$ s.

Why

Adding the three displacements is legitimate because they all point the same way. Had one phase been a descent, its area would have entered with a minus sign, and the distance travelled would then have differed from the height climbed.

Step 5

Average velocity: $v_{av} = \dfrac{15.6}{9.0} = 1.7$ m/s.

Why

The definition, not the midpoint formula: $\dfrac{v_{i} + v_{f}}{2}$ would give $0$ m/s here, since the lift starts and ends at rest, and that absurdity is the standard way of losing this last part.

The conclusion, written out

“Taking upwards as positive, the lift reaches $2.4$ m/s, slows down at $-0.80$ m/s² during the last phase, climbs $15.6$ m in all and travels at an average velocity of $1.7$ m/s.”

The classic mistake on this problem: The classic mistake is to apply one equation of motion to the whole $9.0$ s with some average acceleration. The acceleration is constant on each phase and on no larger interval, so the four equations are simply not available over the nine seconds, while the areas always are.

Learn by heart

• Write the axis in words before the first number. It is one line and it makes every sign on the page verifiable.
• The four equations are chosen by the quantity that is MISSING, and all four demand a constant acceleration.
• Same signs for $v$ and $a$ : the speed grows. Opposite signs: the speed falls. The direction of travel is read on $v$ alone.
• In free fall the magnitude is always $9.8$ m/s²; the sign comes from your axis, $-9.8$ with an upward axis, $+9.8$ with a downward one.
• Slope of a position-time graph gives a velocity, slope of a velocity-time graph gives an acceleration, area under a velocity-time graph gives a displacement.
• Instantaneous velocity is the slope of a TANGENT, read between two points of the tangent itself, never a position divided by a time.
• With a constant acceleration, the average velocity over an interval equals the instantaneous velocity at its MIDPOINT: this replaces a tangent nobody drew.
• A speed is never negative, and a closed trip has a zero average velocity whatever distance was covered.
• When a motor stops, the acceleration changes at once and the velocity does not: the velocity at the end of one phase starts the next.
• A letter answer passes three checks before any number: its unit, a limit case you already know, and the way it moves when one data value grows.

Frequently asked questions

What is the difference between distance travelled and displacement?

The displacement is the single arrow from the starting point to the finishing point, so it carries a sign and it can be zero. The distance travelled adds up the ground covered leg by leg and is always positive. A walk of thirty metres out and thirty metres back has a displacement of zero and a distance travelled of sixty metres, and exam questions choose between the two words deliberately.

Does a negative acceleration mean the object is slowing down?

Not by itself. The sign of the acceleration says in which direction the velocity is being changed along the axis you chose. Compare it with the sign of the velocity: if the two agree, the speed grows, and if they disagree, the speed falls. An object moving backwards with a negative acceleration is speeding up, not slowing down.

How do I find an instantaneous velocity without calculus?

Read the slope of the tangent to the position-time graph at that instant, taking two points that both lie on the tangent line and are far apart. If no tangent is drawn and the acceleration is constant, use the other route: the average velocity over a short interval centred on that instant is exactly the instantaneous velocity at its midpoint.

Is g positive or negative in a free fall problem?

Both, depending on the axis you decide to use, which is why it has to be written down. The magnitude is nine point eight metres per second squared and never changes. With the upward direction taken as positive the acceleration is minus that value, and with the downward direction taken as positive it is plus that value, which often makes depth problems simpler.

Why is the area under a velocity graph a displacement and not a distance?

Because a region lying below the time axis corresponds to a piece of motion in the negative direction, and it therefore counts negatively. Adding the areas with their signs gives the displacement, while adding them without their signs gives the distance travelled. The two answers agree only when the velocity keeps one sign over the whole interval.

Does a drone or a rocket start falling the moment its motors stop?

No. When the motors stop, only the acceleration changes at once: the velocity keeps the value it had, so an object that was climbing goes on climbing, more and more slowly, until its velocity reaches zero. That is the true top. It then comes back down past the altitude where the motors stopped with the same speed it had there, now directed downwards.

Practise it

Corrected exercises: Motion in a straight line, PHYS 101

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.

10 corrected exercises
100 points
150 minutes

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-kinematics-1d. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: motion in a straight line (PHYS 101)

The essentials

On your formula sheet: this chapter's lines, and when each one holds

The four equations, sorted by what is MISSING

The sign of a tells you nothing about speeding up

Free fall: one acceleration for the whole flight

The rules in table form

Which equation, according to the quantity that is missing

The mistakes that cost marks

1. Reading a negative acceleration as a reverse gear

2. Reporting the area under a velocity-time graph as the distance travelled

3. Averaging two velocities instead of dividing the displacement by the time

4. Using g=−9.8g = -9.8g=−9.8 m/s² with an axis that points downwards

5. Concluding that the acceleration is zero at the top of a throw

6. Using the midpoint formula when the acceleration is not constant

7. Handing in a negative velocity as a speed

8. Dividing one position by one time to get an instantaneous velocity

9. Letting friction turn a motion around

10. Taking the instant the motors fail as the top of the climb

Which method to choose

Which equation of motion to write down

What to do with a graph: slope or area

How the answer is expected to be written

Solving a problem of uniformly accelerated motion

Two objects that meet

A motion in two phases

Check before you hand in

The typical problem, taken apart

A hospital lift, in three phases

Learn by heart

Frequently asked questions

What is the difference between distance travelled and displacement?

Does a negative acceleration mean the object is slowing down?

How do I find an instantaneous velocity without calculus?

Is g positive or negative in a free fall problem?

Why is the area under a velocity graph a displacement and not a distance?

Does a drone or a rocket start falling the moment its motors stop?

Corrected exercises: Motion in a straight line, PHYS 101

See also

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4. Using $g = -9.8$ m/s² with an axis that points downwards