Revision sheet: kinematics in one and two dimensions (203-SN1-RE)
Kinematics in 203-SN1-RE (formerly 203-NYA) is not the kinematics of high school: the course is built on differential calculus, so v is the derivative of x, a the derivative of v, and the way back goes through an integral with its constant. The four constant-acceleration equations are no longer the subject of the chapter, they are a special case of it, the one where a is constant.
This sheet does not repeat the course. It gathers what costs marks on this particular chapter: the signs, the formulas applied outside their conditions, and the two motions of a projectile that students link badly. Each trap is written the way it appears on a student paper, with what it costs. It is written for CEGEP students in the Science program, and the corrected exercise set on the same chapter puts every reflex to the test.
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The thread of the chapter
No motion exists without an axis. As long as the axis and its origin are not written on the page, the sign of g, the sign of v and the sign of Δx float freely, and that is where half of the marks of the first 203-SN1-RE exam are lost.
The three definitions everything else follows from
•v(t)=dtdx and a(t)=dtdv. These are definitions, valid always, even when a varies.
•The way back is an integral: v(t)=v0+∫0tadt′ and x(t)=x0+∫0tvdt′. The constants v0 and x0 are not decoration, they carry the initial state.
•On an x(t) graph, the SLOPE gives v. On a v(t) graph, the slope gives a and the SIGNED AREA gives Δx.
•Displacement Δx=xf−xi: a vector, which can be zero after a long trip. Distance travelled: the sum of the lengths, always positive.
•Average velocity =ΔtΔx. Average speed =Δtdistance. The two coincide only if the motion never reverses.
On the left, the slope of the tangent to x(t) gives v at that instant; on the right, the area under v(t) between two instants gives the displacement Δx.
In 203-SN1-RE, the question “which equation should I use” only makes sense if a is constant. If the problem gives a(t) as an expression, the only answer is: integrate.
The four constant-acceleration equations, and their condition of use
•vf=vi+aΔt (no Δx)
•Δx=viΔt+21aΔt2 (no vf)
•vf2=vi2+2aΔx (no Δt)
•Δx=2vi+vfΔt (no a)
•ONE condition, and it is not negotiable: a constant over the whole interval considered. A change of phase (motor cut off, contact with the ground, a parachute opening) starts a NEW interval and a new set of initial values.
Five quantities, four equations: each one is the one that does without the fifth. Spot the quantity that is neither given nor asked for and the equation picks itself.
The sign, the only real difficulty of the chapter
•The axis is a choice, but it is SINGLE for the whole problem. Once the axis points up, g enters the equations as a=−9.8 m/s2, including on the way up.
•“Slowing down” does not mean a<0. Slowing down means v and a of OPPOSITE SIGNS. Speeding up means v and a of the same sign.
•An object can have a constant, negative a while having v>0 then v<0: it slows down, stops, then sets off backward while speeding up.
•At the top of a vertical throw, v=0 but a=−9.8 m/s2: nothing vanishes in the acceleration, otherwise the object would hang there.
•In 2D you do not choose one axis but two, and the same sign of g holds for the whole path, the rise included.
One acceleration, constant and negative (the single slope of the line): before t=4 s the object slows down, after it moves backward while speeding up. The sign of a alone never says whether the object speeds up.
Writing “axis Oy pointing up, origin at the ground, t=0 at the throw” before the first equation is often worth half a mark of rigour and prevents three quarters of the errors that follow.
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.Setting the acceleration to zero at the top of a projectile's path or of a vertical throw
the whole question, 5 marks
What not to write
“At the highest point the velocity is zero, so the acceleration is too, so vy stays zero afterward.”
What to write
“At the top, vy=0 but ay=−9.8 m/s2, as everywhere else: it is precisely this acceleration that brings the object back down.”
The three acceleration arrows have the same length and direction, the top included. At the top, only the velocity arrow has changed: it has become horizontal, it has not disappeared.
Why: Velocity and acceleration are two independent quantities at a given instant. If a vanished at the top, the object would stay there motionless forever, which nobody has ever observed.
2.Mixing up the signs in free fall by taking g positive on an upward axis
3 marks, and an impossible fall time
What not to write
“Thrown upward at 15 m/s from 20 m: y=20+15t+4.9t2.”
What to write
“Axis pointing up, so a=−9.8 m/s2 and y=20+15t−4.9t2. The height must end up decreasing, not exploding.”
Why: The sign of g is not a property of gravity, it is the projection of gravity onto the axis you have just chosen. Choosing the axis upward and writing +g amounts to making the object fall toward the sky.
3.Applying the four constant-acceleration equations to an acceleration that varies
the whole question, 6 marks
What not to write
“a(t)=6t m/s2, vi=0 and Δt=4 s, so Δx=21aΔt2=21(6)(16)=48 m.”
What to write
“a depends on time, so none of the four equations applies. Integrate: v(t)=3t2, then x(t)=t3, so Δx=43=64 m.”
Why: The four equations are the result of an integration done once and for all WITH a constant. Outside that assumption they are not approximate, they are wrong.
4.Losing the constant of integration on the way up from a to v then to x
2 marks per integration, often 4 marks on the question
What not to write
“a(t)=4, so v(t)=4t and x(t)=2t2.”
What to write
“a(t)=4, so v(t)=4t+v0 and x(t)=2t2+v0t+x0, with v0 and x0 read from the problem.”
Why: An antiderivative is never unique. Writing v(t)=4t silently decides that the object started from rest, which the problem almost never says.
5.Confusing distance travelled with displacement when the motion changes direction
2 marks, and a wrong average speed in the same stroke
What not to write
“The train moves forward for 6 s at 10 m/s then backs up for 4 s at 5 m/s: the distance travelled is 60−20=40 m.”
What to write
“The displacement is 60−20=40 m, but the distance travelled is 60+20=80 m. The area under v(t) counts the regions below the axis as NEGATIVE for the displacement, as POSITIVE for the distance.”
Why: Displacement is a difference of positions, distance a sum of lengths. The question “what distance did it travel” never asks for the first.
6.Computing an average velocity as the arithmetic mean of two velocities outside constant acceleration
2 marks, and an answer that is always too large
What not to write
“40 km at 80 km/h then 40 km at 120 km/h: average velocity =280+120=100 km/h.”
What to write
“Average velocity =total timetotal distance=0.5+3180=96 km/h. The mean of vi and vf is valid ONLY if a is constant.”
Why: 2vi+vf is one of the four constant-acceleration equations: it therefore inherits their condition. Over two legs travelled at constant velocity, it is TIME that weights them, and more time is spent on the slow leg.
7.Using the range formula on a throw that does not land at its starting height
the whole question, 6 marks
What not to write
“Thrown from the top of a 25 m roof at 20 m/s and 53°: range =gv02sin2θ=9.8400×0.9614=39.2 m.”
What to write
“The range formula assumes a launch and a landing at the SAME height. Here we solve 0=25+15.97t−4.9t2, so t=4.42 s and x=12.04×4.42=53.1 m.”
Why: The range formula is a shortcut proved for the symmetric case. As soon as a height appears in the problem, it underestimates the result, here by 14 m, a quarter of the answer.
8.Adding relative velocities as magnitudes instead of as vectors
3 marks, plus the uncomputed drift
What not to write
“The swimmer goes at 2 m/s, the current at 1.5 m/s: the velocity relative to the bank is 3.5 m/s.”
What to write
“The swimmer aims perpendicular to the bank, so vs/bank=vs/water+vwater/bank component by component: v=22+1.52=2.5 m/s, deflected 36.9° downstream.”
Swimmer relative to the water (teal) and current (grey) are perpendicular: placed tip to tail, they give the velocity relative to the bank (black), 22+1.52=2.5 m/s at 36.9°, not 3.5 m/s.
Why: Velocities add as scalars only if they are collinear and in the same direction. In a river crossing they are perpendicular, so it is Pythagoras, not a sum.
9.Declaring the acceleration zero in uniform circular motion
3 marks, and the next chapter becomes incomprehensible
What not to write
“The ball turns at constant speed, so a=0.”
What to write
“The MAGNITUDE of the velocity is constant, but its DIRECTION changes: ac=rv2, directed toward the centre. Only the tangential component is zero.”
On the left, the velocity keeps its magnitude but turns: the acceleration points toward the centre and never vanishes. On the right, the speed increases: at is added and a leaves the radius.
Why: Acceleration is the derivative of a VECTOR. A vector can change without its norm changing, and that is exactly what happens on a circle. Without this acceleration, there is no centripetal force in dynamics.
Which method to choose
Which constant-acceleration equation to choose
List the five quantities vi, vf, a, Δx, Δt, then spot the one that is NEITHER given NOR asked for: the equation to use is the one where it does not appear.
If time is neither given nor asked for → vf2=vi2+2aΔx
Example: braking from 25 m/s to a stop over 50 m: a=−6.25 m/s2
this is the equation for stopping distances and maximum heights
If displacement is neither given nor asked for → vf=vi+aΔt
Example: free fall from rest, v after 3 s: v=−29.4 m/s
If acceleration is neither given nor asked for → Δx=2vi+vfΔt
Example: from 10 to 30 m/s in 4 s: Δx=80 m
If final velocity is neither given nor asked for → Δx=viΔt+21aΔt2
Example: position at any time t, the position equation itself
If two of the five quantities are missing → set up a system with two of the equations, or find the hidden constraint (meeting, same time, same position)
typical case: the chase between two vehicles, where the common unknown is Δt
No branch applies if a is not constant over the interval: in that case you integrate. And if the motion changes regime along the way, split it into intervals, each with its own vi and xi.
Which setup according to the form of the motion
What the problem gives, before any calculation
If the problem gives x(t) or v(t) as an expression → differentiate, and never use the four equations
Example: x(t)=t3−6t2, so v=3t2−12t and a=6t−12
If the problem gives a(t) as an expression → integrate twice, keeping v0 and x0
Example: a=6t with v0=2 gives v=3t2+2
If a is constant and the motion is along a line → the four equations, after setting the axis
Example: free fall, braking, takeoff
If the object is thrown and only gravity acts → two separate motions, uniform in x and constant acceleration in y, linked by the ONE common time
Example: x=v0cosθt and y=y0+v0sinθt−21gt2
If two reference frames move relative to each other → vector addition vA/C=vA/B+vB/C, component by component
Example: swimmer and current, plane and wind, passenger and moving walkway
If the path is a circle travelled at constant speed → a=rv2 toward the centre, and nothing else
Example: v=6 m/s on r=4 m gives ac=9 m/s2
The word “projectile” almost never appears in a problem: you recognize it from “thrown”, “kicked”, “leaves the table”, “fired”, followed by the absence of any force other than weight.
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 projectile problem
When to use it: An object is thrown, kicked or leaves an edge, and nothing but gravity acts afterward.
1Set the frame: origin at the launch point or at the ground, axis Ox horizontal in the direction of motion, axis Oy upward. Write it down, do not just think it.
2Resolve the initial velocity: v0x=v0cosθ and v0y=v0sinθ, stating the unit and checking that the calculator is in degrees.
3Write the two position equations: x(t)=x0+v0xt (uniform motion, since ax=0) and y(t)=y0+v0yt−21gt2.
4Translate the question into a CONDITION on x or on y: “hits the ground” gives y=0, “at the top” gives vy=0, “clears the wall” gives x=d.
5Solve this condition for t, keeping the positive root and justifying the rejection of the other.
6Substitute this t into the other equation to get the quantity asked for, then conclude with the unit.
Concluding sentence
“With the Oy axis pointing up and the origin at the ground, the condition y(t)=0 gives t=4.42 s, the only positive root; substituting into x(t), the projectile hits the ground 53.1 m from the foot of the building.”
The trap: Step 4. Many papers compute everything they know how to compute without ever writing the condition that translates the question, and end up with five correct results, none of which is the answer.
Marking: Typically 1 mark for the frame, 2 for the velocity components, 2 for the two position equations, 2 for the condition and its solution, 1 for the conclusion and the unit.
Going back from acceleration to position
When to use it: The problem gives a(t) as an expression, plus the initial conditions.
1Write v(t)=∫a(t)dt showing the constant: v(t)=…+C1.
2Determine C1 with the initial condition on the velocity, writing the equality v(0)=v0 explicitly.
3Write x(t)=∫v(t)dt=…+C2, then determine C2 with x(0)=x0.
4Answer the question asked with the function you obtained, and check that x(0) and v(0) give back the data of the problem.
Concluding sentence
“Since a(t)=6t, we have v(t)=3t2+C1; the condition v(0)=2 m/s gives C1=2, so v(t)=3t2+2, then x(t)=t3+2t+C2 with x(0)=0, hence x(t)=t3+2t.”
The trap: Determining C2 while forgetting that the term C1t has just been added: the constant of the first integration becomes a linear term in the second, it does not disappear.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The sign of the result against the declared axis
Reread the axis set in step 1, then look at the sign of each answer. A positive final velocity for a falling object, with the axis pointing up, is certainly wrong.
A vf=+27.3 m/s for a ball hitting the ground, axis pointing up: it is −27.3 m/s, and the magnitude alone is 27.3 m/s.
The time substituted into the OTHER equation
A projectile always gives two equations. The t found with one must give, in the other, a physically possible value: a positive height, a range smaller than the one in the problem.
A t that gives y=−12 m when the ground is at y=0 signals a negative root kept by mistake.
The order of magnitude, against a known case
Compare with memorized benchmarks: a sprinter at 10 m/s, a car in town at 14 m/s, a one-storey fall in 0.8 s. An answer ten times too large shows without recalculating.
A range of 530 m for a ball thrown at 20 m/s: that is a factor of 10, almost always a misplaced decimal point or a misplaced g.
The calculator mode and the unit of the angle
Before any projectile, check DEG on the screen. A cos53 in radians is −0.918 instead of 0.602, and the negative sign sends the projectile backward.
A negative horizontal component with an angle given between 0 and 90°: the calculator is in radians.
The table of values for a motion given by an expression
Enter x(t) in the calculator's function menu, display the table around the time you found, and check that the position changes direction where v changes sign.
A table that keeps increasing after the announced turnaround time says that the derivative was computed wrongly.
The typical problem, taken apart
A throw from the roof of a building, from the frame to the impact velocity
From the roof of a 25.0 m building, a ball is thrown with an initial velocity of 20.0 m/s directed 53.0° above the horizontal.
Find the time of flight, the horizontal distance at the point of impact and the magnitude of the impact velocity. Take g=9.80 m/s2 and neglect air resistance.
Launch and landing are not at the same height: the path passes back through the roof level (dashed) well before reaching the ground, and that is all that rules out the range formula.
Step 1
Frame: origin at the foot of the building, axis Ox horizontal in the direction of the throw, axis Oy upward. So x0=0, y0=25.0 m and ay=−9.80 m/s2.
Why
This is the step that fixes every sign in the question. Written down, it makes the −9.80 of the following lines indisputable; left out, it lets the marker decide that the next sign was luck.
Step 2
v0x=20.0cos53.0°=12.04 m/s and v0y=20.0sin53.0°=15.97 m/s.
Why
A projectile is never treated with v0 as it stands: the two axes do not have the same acceleration, so they do not have the same equation. The resolution is what separates the two problems.
Step 3
x(t)=12.04t and y(t)=25.0+15.97t−4.90t2.
Why
The t is the same in both lines: it is the only link between the horizontal and the vertical. The rest of the question consists in finding it on one side to use it on the other.
Step 4
Impact condition: y(t)=0, that is 4.90t2−15.97t−25.0=0, so t=9.8015.97+255.0+490.0=4.42 s. The negative root −1.16 s is rejected, since the ball does not exist before the throw.
Why
This is where the question becomes an equation. Rejecting the negative root takes one sentence and earns the rigour mark; without the sentence, the marker sees two answers and counts neither.
Step 5
x(4.42)=12.04×4.42=53.1 m.
Why
The time found on the vertical axis is carried over to the horizontal axis: it is the only way to get the range when launch and landing are not at the same height.
Step 6
vy(4.42)=15.97−9.80×4.42=−27.3 m/s, so v=12.042+27.32=29.8 m/s.
Why
The horizontal component has not changed since the throw, since ax=0. Only the vertical one evolves, and its negative sign says the ball is going down at the moment of impact.
Step 7
Check with the speed-height relation: v2=v02+2gh=400+2(9.80)(25.0)=890, so v=29.8 m/s.
Why
This relation involves neither the time nor the angle: it checks the impact speed by a path completely independent of the previous one. Two agreeing methods are worth more than a recheck of the arithmetic.
The conclusion, written out
“The ball stays in the air for 4.42 s, hits the ground 53.1 m from the foot of the building, and arrives at 29.8 m/s.”
The classic mistake on this problem: Applying R=gv02sin2θ and answering 39.2 m. The formula is right, but it is proved for a return to the starting height: here it gives the distance at the moment the ball passes back through roof level, not at the ground.
Learn by heart
•v=dtdx and a=dtdv are true ALWAYS; the four constant-acceleration equations are true only if a is CONSTANT.
•On a v(t) graph: the slope is a, the signed area is Δx, the area in absolute value is the distance travelled.
•Slowing down means v and a of OPPOSITE signs. The sign of a alone says nothing.
•At the top of a projectile's path: vy=0, vx unchanged, a=−g as everywhere else.
•A projectile is two motions and ONE time: x=v0cosθt and y=y0+v0sinθt−21gt2.
•R=gv02sin2θ requires a launch and a landing at the SAME height, and so does the optimal angle of 45°.
•Relative velocities: vA/C=vA/B+vB/C, component by component, never as magnitudes.
•Uniform circular motion: v constant in magnitude, ac=rv2 toward the centre, never zero.
Frequently asked questions
Why is the acceleration not zero at the top of a projectile's path?
Because velocity and acceleration are two distinct quantities. At the top, the vertical component of the velocity vanishes for an instant, but gravity acts without interruption: the acceleration is 9.8 metres per second squared downward throughout the flight. It is precisely what brings the object back down. If it vanished at the top, the object would hang there.
Which constant-acceleration equation should I choose in a problem?
List the five quantities involved: initial velocity, final velocity, acceleration, displacement, time. Spot the one that is neither given in the problem nor asked for by the question. The equation to use is the one where that quantity does not appear. This reading avoids pointless systems and takes five seconds.
When can the range formula of a projectile be used?
Only when the projectile lands at exactly the height it left from, for example a ball kicked from the ground and caught at ground level. As soon as the problem mentions a cliff, a roof, a table or a raised basket, the formula underestimates the distance, and you have to go back to the condition on the height, then substitute the time you find.
How do I set the sign of g in a free fall problem?
The sign of g depends on the axis you choose, not on gravity. Axis pointing up: the acceleration is minus 9.8 throughout the motion, the rise included. Axis pointing down: it is plus 9.8. The only forbidden move is to change axis in the middle of a problem or to use a different sign depending on whether the object is going up or down.
What is the difference between distance travelled and displacement?
Displacement is the difference between the final position and the initial position: it has a sign and can be zero after a round trip. Distance travelled adds up the lengths of all the legs and is always positive. On a velocity-time graph, the first counts areas below the axis as negative, the second as positive.
How do I handle a relative velocity problem such as a river crossing?
You add vectors, never numbers. The velocity relative to the bank is the vector sum of the swimmer's velocity relative to the water and the velocity of the current. Project onto two axes, add component by component, then go back to a magnitude and an angle. The width of the river is handled with the perpendicular component alone.
Practise it
Corrected exercises: Kinematics in one and two dimensions, 203-SN1-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.
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Get in touch for a first session. In kinematics we always start with the same thing: set the axis, write the origin, and write no equation until both are on the page.
