Kinematics and forces: A-level Maths OCR revision notes

3.02

Kinematics graphs and suvat

Displacement, velocity and acceleration 3.02

Definitions

Displacement: the vector from a fixed origin to the object's position, measured in metres ( $m$ ).
Distance travelled: the scalar length of the path followed, which is never less than the magnitude of the displacement.
Velocity: the rate of change of displacement, a vector measured in $m s^{- 1}$ .
Speed: the magnitude of the velocity vector, a scalar which is never negative.
Acceleration: the rate of change of velocity, a vector measured in $m s^{- 2}$ .

Key results

In one dimension a sign convention (say, rightward or upward as positive) turns each vector into a signed scalar, so a negative velocity simply means motion in the negative direction.
Deceleration means the acceleration acts opposite to the direction of motion; whether the number $a$ is negative depends entirely on which direction was chosen as positive.

Notes

A change of direction at constant speed is still an acceleration: acceleration is a change in the velocity vector, and direction is part of that vector.
Distance and displacement agree only while the motion stays in one direction — an object that returns to its starting point has zero displacement but a non-zero distance travelled.
State the positive direction explicitly at the start of a solution; almost every sign error in mechanics traces back to not having done so.
Speed is never negative, but velocity can be — a 'speed' of $- 3 m s^{- 1}$ is a nonsensical quantity, and signals the vector/scalar distinction has been lost somewhere in the working.

Motion graphs 3.02

Key results

On a displacement-time graph the gradient at a point gives the velocity at that instant, and a horizontal section means the object is at rest.
On a velocity-time graph the gradient gives the acceleration, and the area between the graph and the time axis gives the displacement.
Area below the time axis on a velocity-time graph counts as negative displacement, so total distance is found by adding the areas as positive amounts while displacement adds them with sign.
Constant acceleration appears as a straight line on a velocity-time graph, so the area is a trapezium of parallel sides $u$ and $v$ and width $t$ , which is exactly $s = \frac{1}{2} (u + v) t$ .

Notes

Reading a graph is often faster and less error-prone than reaching immediately for a formula, particularly for a journey made of several stages with different accelerations.
Checking units confirms the interpretation of an area: $m s^{- 1}$ multiplied by $s$ gives $m$ , so the area under a velocity-time graph must be a displacement.
A displacement-time graph can never have a vertical section (that would mean infinite speed) — a kink is fine, a vertical jump signals a plotting error.

The suvat equations 3.02

Definitions

suvat variables: $s$ displacement ( $m$ ), $u$ initial velocity ( $m s^{- 1}$ ), $v$ final velocity ( $m s^{- 1}$ ), $a$ acceleration ( $m s^{- 2}$ ) and $t$ time ( $s$ ).

Key results

$v = u + a t$ — the only equation without $s$ .
$s = u t + \frac{1}{2} a t^{2}$ — the only equation without $v$ .
$v^{2} = u^{2} + 2 a s$ — the only equation without $t$ .
$s = \frac{1}{2} (u + v) t$ — the only equation without $a$ .
A fifth form, $s = v t - \frac{1}{2} a t^{2}$ , follows by eliminating $u$ , and is occasionally quicker when the final velocity is known but the initial one is not.

Method

List the five variables and fill in the three that are given, with signs consistent with the chosen positive direction.
Identify the one variable required, and hence the one variable that is neither given nor required.
Select the equation that omits that unused variable, substitute, and solve.
State the answer with its unit, and sanity-check the sign against the chosen positive direction.

Notes

Every suvat equation assumes the acceleration is constant throughout the interval; a journey in stages with different accelerations must be split into one suvat calculation per stage.
Where the acceleration varies, calculus replaces suvat: velocity is $\frac{d s}{d t}$ , acceleration is $\frac{d v}{d t}$ , and displacement is recovered by integrating velocity with respect to time.
Mixing units — a speed in $km h^{- 1}$ with a time in seconds, say — is a frequent source of error; convert everything to metres and seconds before substituting.
Writing all five known/unknown values down before choosing an equation, rather than picking an equation first, avoids the common trap of selecting one that still contains an unknown you haven't found yet.

Vertical motion under gravity 3.02

Key results

A body moving freely under gravity has constant acceleration of magnitude $g = 9.8 m s^{- 2}$ directed vertically downward, so the suvat equations apply exactly.
At the highest point of a vertical throw the velocity is instantaneously zero, but the acceleration is still $9.8 m s^{- 2}$ downward.
For a throw returning to its launch height the motion is symmetric: the time up equals the time down, and the speed on return equals the speed of projection.

Method

Choose and state a positive direction, conventionally upward.
Write $u$ , $a$ and $s$ with signs matching that choice, so that $a = - 9.8 m s^{- 2}$ when upward is positive.
Apply the appropriate suvat equation, reading a negative displacement as a position below the starting point.
Interpret the answer physically before rounding to 3 significant figures.

Notes

When a body lands below its launch point, $s = u t + \frac{1}{2} a t^{2}$ becomes a quadratic in $t$ with one positive and one negative root; discard the negative root as unphysical.
The standard model treats the body as a particle and neglects air resistance; both assumptions should be stated whenever a question asks about the modelling.
Take $g$ as either $9.8 m s^{- 2}$ or $10 m s^{- 2}$ exactly as the question specifies — mixing the two values within one calculation gives an inconsistent answer.

Worked examples

Worked example

A car travelling at $30 m s^{- 1}$ brakes uniformly and covers $90 m$ while slowing to $12 m s^{- 1}$ .

Find the acceleration and the time taken.

Show worked solution

Known: $u = 30 m s^{- 1}$ , $v = 12 m s^{- 1}$ , $s = 90 m$ ; required $a$ , with $t$ unused, so use $v^{2} = u^{2} + 2 a s$ .

Then:

144 = 900 + 180 a

so:

a = \frac{144 - 900}{180} = \frac{- 756}{180} = - 4.2 m s^{- 2}

a deceleration of $4.2 m s^{- 2}$ .

For the time use $v = u + a t$ :

12 = 30 - 4.2 t

so:

t = \frac{18}{4.2} = 4.29 s

(3 s.f.).

Check with:

s = \frac{1}{2} (u + v) t

\frac{1}{2} (30 + 12) (4.2857) = 21 \times 4.2857 = 90 m

as given.

Worked example

A ball is thrown vertically upward at $20 m s^{- 1}$ .

Taking $g = 9.8 m s^{- 2}$ , find the maximum height reached and the total time for the ball to return to its starting point.

Show worked solution

At maximum height, $v = 0$ .

Using $v^{2} = u^{2} - 2 g s$ (taking upward as positive, so $a = - g$ ):

0 = 2 0^{2} - 2 (9.8) s

giving:

s = \frac{400}{19.6} = 20.4 m

(3 s.f.).

Time to reach maximum height, using $v = u - g t$ : $0 = 20 - 9.8 t$ , so:

t = \frac{20}{9.8} = 2.04 s

By symmetry of the motion, the time to fall back to the starting height equals the time to rise, so the total time is:

2 \times 2.04 = 4.08 s

(3 s.f.).

Worked example

A train starts from rest and accelerates uniformly at $0.5 m s^{- 2}$ for $40 s$ , then travels at constant speed for $3$ minutes, then decelerates uniformly to rest in a further $25 s$ .

Sketch the velocity-time graph and find the total distance travelled and the final deceleration.

Show worked solution

The graph is a trapezium: a straight line rising from the origin, a horizontal section, then a straight line falling to the time axis.

Cruising speed:

v = u + a t = 0 + 0.5 (40) = 20 m s^{- 1}

Stage 1 area (triangle):

\frac{1}{2} (40) (20) = 400 m

Stage 2 area (rectangle), with $3$ minutes $= 180 s$ :

20 \times 180 = 3600 m

Stage 3 area (triangle):

\frac{1}{2} (25) (20) = 250 m

Total distance:

= 400 + 3600 + 250 = 4250 m = 4.25 km

over a total time of $245 s$ .

Deceleration: the gradient of the final section is:

\frac{0 - 20}{25} = - 0.8

a deceleration of $0.8 m s^{- 2}$ .

Worked example

A stone is thrown vertically upward at $14.7 m s^{- 1}$ from the top of a cliff and lands on the beach $49 m$ below the point of projection.

Taking $g = 9.8 m s^{- 2}$ , find the time of flight and the speed with which the stone lands.

Show worked solution

Take upward as positive:

u = 14.7 m s^{- 1}

a = - 9.8 m s^{- 2}

$s = - 49 m$ .

Using:

s = u t + \frac{1}{2} a t^{2}

- 49 = 14.7 t - 4.9 t^{2}

so:

4.9 t^{2} - 14.7 t - 49 = 0

Dividing by $4.9$ :

t^{2} - 3 t - 10 = 0

which factorises as:

(t - 5) (t + 2) = 0

giving $t = 5 s$ (rejecting $t = - 2 s$ as unphysical).

Landing velocity from $v = u + a t$ :

v = 14.7 - 9.8 (5) = - 34.3 m s^{- 1}

so the stone lands at a speed of $34.3 m s^{- 1}$ , moving downward.

Check with $v^{2} = u^{2} + 2 a s$ :

14. 7^{2} + 2 (- 9.8) (- 49) = 216.09 + 960.4 = 1176.49

and:

1176.49 = 34.3

confirming the result.

3.02

Vectors in kinematics

Motion in two dimensions with vectors 3.02

Key results

The suvat equations hold as vector equations, $v = u + a t$ and $s = u t + \frac{1}{2} a t^{2}$ , and so may be applied to each component separately.
Where the acceleration has constant, separate horizontal and vertical components, the two directions are completely independent and share only the time $t$ — this is exactly the structure of projectile motion, covered in Further mechanics.
For $v = v_{1} i + v_{2} j$ , the speed is $∣ v ∣ = v_{1}^{2} + v_{2}^{2}$ and the direction is $arctan (\frac{v _{2}}{v _{1}})$ measured from the $i$ direction.

Notes

Position, displacement, velocity and acceleration are all vectors; distance, speed and time are the scalars, and only scalars may be added arithmetically.
A vector answer is not finished until both a magnitude and a direction have been given, unless the question asks only for one of them.
Component equations for perpendicular directions can be solved completely independently — resist the urge to combine the $i$ and $j$ equations together, since they share nothing but the time $t$ .

Worked example

A particle has initial velocity $(3 i - 2 j) m s^{- 1}$ and moves with constant acceleration $(- i + 4 j) m s^{- 2}$ .

Find its speed after $5 s$ and its distance from the starting point at that moment.

Show worked solution

Velocity:

v = u + a t = (3 i - 2 j) + 5 (- i + 4 j) = - 2 i + 18 j

so the speed is:

(- 2)^{2} + 1 8^{2} = 328 = 18.1 m s^{- 1}

(3 s.f.).

Displacement:

s = u t + \frac{1}{2} a t^{2} = 5 (3 i - 2 j) + \frac{1}{2} (25) (- i + 4 j) = (15 i - 10 j) + (- 12.5 i + 50 j) = 2.5 i + 40 j

The distance is therefore:

2. 5^{2} + 4 0^{2} = 1606.25 = 40.1 m

(3 s.f.).

Note the particle has moved a long way in the $j$ direction and barely at all in the $i$ direction, since its initial $i$ velocity has been almost entirely reversed.

3.03

Newton's laws and connected particles

Newton's laws of motion 3.03

Definitions

Newton's first law: a body remains at rest or continues at constant velocity unless acted on by a resultant force.
Newton's second law: the resultant force on a body equals its mass times its acceleration, $F = ma$ , with $F$ in newtons ( $N$ ), $m$ in $kg$ and $a$ in $m s^{- 2}$ .
Newton's third law: forces occur in equal and opposite pairs acting on different bodies.
Weight: the gravitational force on a mass, $W = m g$ , so a mass of $2 kg$ has weight $2 \times 9.8 = 19.6 N$ .

Key results

One newton is the resultant force giving a mass of $1 kg$ an acceleration of $1 m s^{- 2}$ , so $1 N = 1 kg m s^{- 2}$ .
It is the resultant force, not any single applied force, that determines the acceleration.
Perpendicular to the direction of motion the acceleration is zero, so the forces in that direction must balance — this is how the normal reaction is usually found.

Method

Draw the body and mark every force acting on it: weight, normal reaction, tension or thrust, friction, and any applied force.
Choose two perpendicular directions, usually along and perpendicular to the acceleration.
Resolve every force into those directions and write $F = ma$ in each, taking the direction of the acceleration as positive.
Solve the resulting equations, then check that each answer has a sensible magnitude and sign.

Notes

Third-law pairs act on different bodies, so they never cancel within the equation of motion of a single body; the weight of a block and the reaction of the table on it are not a third-law pair, since both act on the block.
Standard modelling vocabulary: a particle has negligible size, light means of negligible mass, smooth means frictionless, and inextensible means of fixed length.
Draw the force diagram BEFORE writing any equation — attempting to resolve forces mentally without a labelled sketch is the most common source of a missing or double-counted force.

Connected particles 3.03

Key results

A light inextensible string passing over a smooth pulley has the same tension throughout, and the two particles have accelerations of equal magnitude, though possibly in different directions depending on the geometry.
The force exerted on the pulley is the vector sum of the two tensions pulling on it; for two vertical sections of string this is $2 T$ directed downward.
A rod, unlike a string, can push as well as pull: a negative tension in a rod is a thrust.
Example: $A$ (3 kg) on a smooth table is pulled by $B$ (2 kg) hanging over the edge: $2 g - T = 2 a$ and $T = 3 a$ give $a = \frac{2 g}{5} = 3.92 m s^{- 2}$ and $T = 11.8$ N.
Masses $m_{1} > m_{2}$ hanging either side of a smooth pulley: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ and $T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}$ .

Method

Apply $F = ma$ separately to each particle, taking that particle's own direction of motion as positive.
Add the two equations to eliminate the tension and solve for the acceleration.
Substitute back into either equation to find the tension.
Verify by checking the tension satisfies the other equation as well.

Notes

Treating the two particles as one body is legitimate for finding the acceleration, but the tension is then internal and must not appear in that whole-system equation.
For a pulley at the edge of a table, the hanging particle's weight drives the motion while the particle on the table contributes only its mass (and friction, if the table is rough) — the two accelerations still have equal magnitude.
The direction taken as positive for each particle should match its own actual direction of motion, even though this means the two particles' equations use opposite conventions relative to a fixed external direction (e.g. 'up' for one, 'down' for the other).

Worked example

Particles of mass $5 kg$ and $3 kg$ are joined by a light inextensible string passing over a smooth fixed pulley, and hang vertically.

The system is released from rest.

Taking $g = 9.8 m s^{- 2}$ , find the acceleration, the tension in the string, and the force exerted on the pulley.

Show worked solution

Let the acceleration have magnitude $a$ and the tension be $T$ , the same throughout the string.

For the $5 kg$ particle, taking downward as positive: $5 g - T = 5 a$ .

For the $3 kg$ particle, taking upward as positive: $T - 3 g = 3 a$ .

Adding eliminates $T$ : $2 g = 8 a$ , so:

a = \frac{2 ( 9.8 )}{8} = 2.45 m s^{- 2}

Substituting into the second equation:

T = 3 (9.8) + 3 (2.45) = 29.4 + 7.35 = 36.75 N

so $T = 36.8 N$ (3 s.f.).

Check in the first equation:

5 g - T = 49 - 36.75 = 12.25

and:

5 a = 5 (2.45) = 12.25

which agree.

Both sections of string are vertical and each pulls down on the pulley with force $T$ , so the force on the pulley is $2 T = 73.5 N$ directed vertically downward.

Note that $T$ lies between the two weights $29.4 N$ and $49 N$ , as it must.

3.03

Forces in equilibrium

Forces in equilibrium 3.03

Definitions

Equilibrium of a particle: the resultant of all forces acting on it is zero, so it remains at rest or moves with constant velocity.

Key results

The vector condition $\sum F = 0$ is equivalent to the components summing to zero separately in each of any two perpendicular directions.
Equilibrium is the special case $a = 0$ of Newton's second law, so no separate theory is needed.
Three forces in equilibrium can also be drawn nose-to-tail as a closed triangle, after which the sine or cosine rule gives the unknowns.
A force $F$ at angle $θ$ to the horizontal has components $F cos θ$ horizontally and $F sin θ$ vertically.
Example: a particle of weight 20 N held at rest on a smooth slope at $3 0^{\circ}$ by a force $P$ up the slope: $P = 20 sin 3 0^{\circ} = 10$ N and $R = 20 cos 3 0^{\circ} = 17.3$ N.

Method

Draw a force diagram and choose two convenient perpendicular directions.
Resolve every force into those two directions.
Set the total in each direction equal to zero, giving two simultaneous equations.
Solve for the unknowns, then check the third force direction or a resolved diagonal as a consistency test.

Notes

Choosing the axes well is worth more than any algebraic trick: pick directions that make as many forces as possible lie along an axis.
Equilibrium never means 'no forces act' — it means the forces that do act sum to zero; a particle in equilibrium can still have several substantial forces on it, each balanced by the others.

The inclined plane 3.03

Key results

Resolving the weight of a body on a plane inclined at $θ$ to the horizontal gives $m g sin θ$ down the slope and $m g cos θ$ perpendicular into the slope.
On a smooth plane with no other force perpendicular to the surface, the normal reaction is $R = m g cos θ$ .
A body released on a smooth plane accelerates down the slope at $a = g sin θ$ , independent of its mass.

Notes

Resolving parallel and perpendicular to the slope, rather than horizontally and vertically, is almost always the better choice, since the acceleration and the normal reaction then lie along the axes.
The $sin$ and $cos$ components are easily swapped: check with a limiting case, since at $θ = 0$ the plane is flat and the down-slope component must vanish, which $m g sin 0 = 0$ confirms.
A string parallel to the slope holds a body in equilibrium with tension $T = m g sin θ$ on a smooth plane; a horizontal string instead needs its own components resolved along and perpendicular to the slope.
$a = g sin θ$ being independent of mass is the same fact as heavier and lighter objects falling at the same rate under gravity alone — the incline just reduces the effective acceleration by the factor $sin θ$ .

Worked example

A particle of mass $2 kg$ rests on a smooth plane inclined at $30°$ to the horizontal, held in equilibrium by a light string parallel to the slope.

Taking $g = 9.8 m s^{- 2}$ , find the tension in the string and the normal reaction from the plane.

Show worked solution

Resolving parallel to the slope: the tension balances the component of weight down the slope, so:

T = m g sin 30° = 2 (9.8) (0.5) = 9.8 N

Resolving perpendicular to the slope: the normal reaction balances the component of weight into the slope, so:

R = m g cos 30° = 2 (9.8) (0.866) = 17.0 N

(3 s.f.).