Mini-Lesson

Kinematics

3 missions. Guess, learn, then check. Your place is saved.

Mission 1A phone drops from a 45 m tower. When does it land, and how fast?
Mission 2A sprinter has $v = 3 t - 0.25 t^{2}$ v = 3t − 0.25t². When does she stop accelerating?
Mission 3What angle throws a ball furthest, and why isn't the real answer that?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A phone drops from a 45 m tower. When does it land, and how fast?

Make a guess: A phone drops from a 45 m tower. When does it land, and how fast?

Mechanics · constant acceleration

The language and the graphs

Displacement and velocity have a direction (they can be negative). Distance travelled and speed are their sizes, and are never negative.
On a displacement–time graph, the gradient is the velocity.
On a velocity–time graph, the gradient is the acceleration and the area between the graph and the time axis is the displacement.
Area below the time axis counts as negative displacement. For distance travelled, add the sizes of the areas.

Worked example Mechanics · constant acceleration

Reading a velocity–time graph

Question

A car starts from rest and accelerates uniformly to 12 m/s in 4 s. It travels at 12 m/s for 10 s, then decelerates uniformly to rest in 6 s. Find its acceleration in the first 4 s and the total distance travelled.

1
Gradient of the first section: $a = \frac{12 - 0}{4} = 3$ a = (12 − 0)/4 = 3 m s $^{- 2}$ ⁻²
2
Split the area into a triangle, a rectangle and a triangle: $\frac{1}{2} (4) (12) + 10 (12) + \frac{1}{2} (6) (12)$ 1/2(4)(12) + 10(12) + 1/2(6)(12)
3
Add: $24 + 120 + 36 = 180$ 24 + 120 + 36 = 180

Answer

acceleration 3 m s

^{- 2}

; distance 180 m

⚠️ Watch out

the deceleration is the gradient of the last section,

- 2

m s

^{- 2}

. Do not use 12 m/s for the whole 20 s (that gives 240 m).

Calculate

?A train starts from rest at station A and accelerates uniformly for 20 s to a speed of

V

m/s. It travels at

V

m/s, then decelerates uniformly for 10 s, stopping at station B. The whole journey takes 100 s and A to B is 1,700 m. Sketch the velocity–time graph and find

V

.

Mechanics · constant acceleration

The constant acceleration formulae

On a straight-line velocity–time graph from $u$ u to $v$ v in time $t$ t: gradient $a = \frac{v - u}{t}$ a = (v − u)/t gives $v = u + a t$ v = u + at, and the trapezium area gives $s = \frac{1}{2} (u + v) t$ s = 1/2(u + v)t.
Substituting one into another gives the five suvat formulae:

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

Each formula leaves out one of $s, u, v, a, t$ s, u, v, a, t. List the three you know and the one you want; use the formula without the fifth.
They are valid only when the acceleration is constant.

Quick check

?A cyclist accelerates uniformly from 4 m/s to 10 m/s over a distance of 42 m. Find her acceleration.

Worked example Mechanics · constant acceleration

Stopping distance

Question

A car travelling at 20 m/s brakes with constant deceleration 2.5 m s

^{- 2}

. Find the time it takes to stop and the distance it travels while braking.

1
List the suvat values: $u = 20$ u = 20, $v = 0$ v = 0, $a = - 2.5$ a = −2.5, $t = ?$ t = ?, $s = ?$ s = ?
2
Use $v = u + a t$ v = u + at: $0 = 20 - 2.5 t$ 0 = 20 − 2.5t, so $t = 8$ t = 8
3
Use $v^{2} = u^{2} + 2 a s$ v² = u² + 2as: $0 = 400 - 5 s$ 0 = 400 − 5s, so $s = 80$ s = 80
4
Check with $s = \frac{1}{2} (u + v) t = \frac{1}{2} (20) (8) = 80$ s = 1/2(u + v)t = 1/2(20)(8) = 80

Answer

8 s and 80 m

⚠️ Watch out

a deceleration of 2.5 m s

^{- 2}

means

a = - 2.5

when the direction of motion is positive. Using

+ 2.5

gives a negative time,

t = - 8

.

Mechanics · constant acceleration

Vertical motion under gravity

Modelled as a particle with no air resistance, a body moving freely up or down has constant acceleration $g$ g downwards.
Use $g = 9.8$ g = 9.8 m s $^{- 2}$ ⁻² unless told otherwise. CCEA questions state the value of $g$ g to use, normally 9.8. $g$ g is not a universal constant: it depends on location.
Choose a positive direction and stick to it. With up positive, $a = - 9.8$ a = −9.8 and a point below the start has negative $s$ s.
At the greatest height, $v = 0$ v = 0.

Worked example Mechanics · constant acceleration

A ball thrown up from a cliff

Question

A ball is thrown vertically upwards at 14.7 m/s from the top of a cliff 19.6 m above the sea. Find the time it takes to reach the sea.

1
Take up as positive: $u = 14.7$ u = 14.7, $a = - 9.8$ a = −9.8, $s = - 19.6$ s = −19.6 (the sea is below the start)
2
Use $s = u t + \frac{1}{2} a t^{2}$ s = ut + 1/2at²: $- 19.6 = 14.7 t - 4.9 t^{2}$ −19.6 = 14.7t − 4.9t²
3
Rearrange and divide by 4.9: $t^{2} - 3 t - 4 = 0$ t² − 3t − 4 = 0
4
Factorise: $(t - 4) (t + 1) = 0$ (t − 4)(t + 1) = 0, and $t > 0$ t > 0

Answer

4 s

⚠️ Watch out

using

s = + 19.6

gives a quadratic with no real roots. The ball ends up below where it started, so

s

is negative.

Calculate

?A stone is thrown vertically upwards from ground level at 21 m/s. Find the length of time for which the stone is at least 10 m above the ground. Give your answer to 3 significant figures.

seconds

Mission check

?A phone drops from a 45 m tower. When does it land, and how fast?

Mission 2

A sprinter has $v = 3 t - 0.25 t^{2}$ v = 3t − 0.25t². When does she stop accelerating?

Make a guess: A sprinter has $v = 3 t - 0.25 t^{2}$ v = 3t − 0.25t². When does she stop accelerating?

Mechanics · variable acceleration

Differentiate to go from displacement to acceleration

When acceleration changes with time, suvat does not apply. Use calculus.

v = \frac{d s}{d t} a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}

Instantaneous rest: $v = 0$ v = 0. Greatest or least velocity: $a = 0$ a = 0, as for any stationary point.
For CCEA, calculus in kinematics is Unit A2 2 content.

Worked example Mechanics · variable acceleration

When is the particle at rest?

Question

A particle moves in a straight line. At time

t

seconds its displacement from

O

is

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

metres,

t \geq 0

. Find the times when it is at rest and its acceleration at each of those times.

1
Differentiate: $v = 3 t^{2} - 12 t + 9$ v = 3t² − 12t + 9
2
Set $v = 0$ v = 0 and factorise: $3 (t - 1) (t - 3) = 0$ 3(t − 1)(t − 3) = 0, so $t = 1$ t = 1 or $t = 3$ t = 3
3
Differentiate again: $a = 6 t - 12$ a = 6t − 12
4
Substitute: at $t = 1$ t = 1, $a = - 6$ a = −6; at $t = 3$ t = 3, $a = 6$ a = 6

Answer

at rest when

t = 1

s (

a = - 6

m s

^{- 2}

) and

t = 3

s (

a = 6

m s

^{- 2}

)

⚠️ Watch out

"at rest" means

v = 0

, not

s = 0

. Here

s = 0

at

t = 3

only by coincidence.

Quick check

?A particle has velocity

v = 8 t - 2 t^{2}

m/s at time

t

s. Find its acceleration when

t = 3

.

Mechanics · variable acceleration

Integrate to go back up

Reverse the chain by integrating: $v = \int a d t$ v = ∫ a dt and $s = \int v d t$ s = ∫ v dt.
Each integral gives a constant. Find it from the initial conditions (for example "starts from rest at $O$ O" means $v = 0$ v = 0 and $s = 0$ s = 0 when $t = 0$ t = 0).
Displacement between two times is $\int_{t_{1} t_{2}} v d t$ ∫_t₁^t₂ v dt.
Distance travelled: find where $v = 0$ v = 0, integrate each part separately, then add the sizes. Area below the axis cancels area above if you don't.

Worked example Mechanics · variable acceleration

From acceleration to displacement

Question

A particle moves in a straight line with acceleration

a = (6 t - 4)

m s

^{- 2}

. When

t = 0

it is at

O

with velocity 3 m/s. Find its displacement from

O

when

t = 2

.

1
Integrate $a$ a: $v = 3 t^{2} - 4 t + c$ v = 3t² − 4t + c
2
Use $v = 3$ v = 3 when $t = 0$ t = 0: $c = 3$ c = 3, so $v = 3 t^{2} - 4 t + 3$ v = 3t² − 4t + 3
3
Integrate $v$ v: $s = t^{3} - 2 t^{2} + 3 t + k$ s = t³ − 2t² + 3t + k, and $s = 0$ s = 0 when $t = 0$ t = 0 gives $k = 0$ k = 0
4
Substitute $t = 2$ t = 2: $s = 8 - 8 + 6$ s = 8 − 8 + 6

Answer

6 m

⚠️ Watch out

leaving out the first constant (the initial velocity 3) gives

s = 0

at

t = 2

. Every integration needs its own constant.

Calculate

?A particle moves in a straight line with velocity

v = t^{2} - 4 t + 3

m/s, for

0 \leq t \leq 3

. Find the total distance travelled. Give your answer to 3 significant figures.

metres

Calculate

?A particle starts from rest and moves in a straight line with acceleration

a = (4 - t)

m s

^{- 2}

,

0 \leq t \leq 8

. Find its greatest velocity.

m/s

Mechanics · variable acceleration

Calculus with vectors (A2 only)

In two dimensions, position is a vector: $r = f (t) i + g (t) j$ r = f(t)i + g(t)j.
Differentiate or integrate each component separately: $v = r ˙ = f^{'} (t) i + g^{'} (t) j$ v = r = f'(t)i + g'(t)j, $a = v ˙$ a = v.
When integrating, the constant is a vector, such as $c = p i + q j$ c = pi + qj.
Speed is the magnitude of velocity: $∣ v ∣ = v_{12} + v_{22}$ |v| = √(v₁² + v₂²). Column vectors work the same way.

Calculate

?(A2 only) A particle has position vector

r = (t^{3} - 3 t) i + 2 t^{2} j

metres at time

t

seconds. Find its speed when

t = 2

. Give your answer to 3 significant figures.

m/s

Mission check

?A sprinter has

v = 3 t - 0.25 t^{2}

. When does she stop accelerating?

Mission 3

What angle throws a ball furthest, and why isn't the real answer that?

Make a guess: What angle throws a ball furthest, and why isn't the real answer that?

Mechanics · projectiles

Constant acceleration with vectors

With constant acceleration $a$ a in two dimensions, the suvat formulae hold with vectors:

v = u + a t r = u t + \frac{1}{2} a t^{2} r = \frac{1}{2} (u + v) t r = v t - \frac{1}{2} a t^{2}

Here $r$ r is the displacement from the start. Vectors can be in $i, j$ i, j form or column form.
$v^{2} = u^{2} + 2 a s$ v² = u² + 2as is not used in vector form. Apply it to one component at a time.
For CCEA this screen is Unit AS 2 content; the projectile screens that follow are Unit A2 2.
A projectile is a particle moving freely under gravity: $a = - g j = (0 - g)$ a = −gj = ( 0 ; −g ).

Worked example Mechanics · projectiles

Thrown horizontally from a cliff

Question

A ball is thrown horizontally at 12 m/s from the top of a cliff 19.6 m above the sea. Find how far from the foot of the cliff it lands, and its speed and direction as it hits the sea.

1
Vertical (down positive, $u_{y} = 0$ u_y = 0): $19.6 = \frac{1}{2} (9.8) t^{2}$ 19.6 = 1/2(9.8)t², so $t = 2$ t = 2
2
Horizontal (constant speed): $x = 12 \times 2 = 24$ x = 12 × 2 = 24
3
Velocity components at $t = 2$ t = 2: horizontal 12, vertical $9.8 \times 2 = 19.6$ 9.8 × 2 = 19.6 (down)
4
Speed $= 1 2^{2} + 19. 6^{2} = 23.0$ = √(12² + 19.6²) = 23.0; angle below horizontal $= tan^{- 1} \frac{19.6}{12} = 58. 5^{\circ}$ = tan⁻¹19.6/12 = 58.5°

Answer

24 m from the cliff, at 23.0 m/s,

58. 5^{\circ}

below the horizontal

⚠️ Watch out

tan^{- 1} \frac{12}{19.6} = 31. 5^{\circ}

is the angle with the vertical. Say which line your angle is measured from.

Quick check

?A particle starts at the origin with velocity

(2 i - 3 j)

m/s and has constant acceleration

(i + 2 j)

m s

^{- 2}

. Find its position vector after 4 seconds.

Mechanics · projectiles

Time of flight, greatest height and range (A2 only)

Launch at speed $U$ U, angle $α$ α above the horizontal: components $U cos α$ Ucos α (horizontal, constant) and $U sin α$ Usin α (vertical, acceleration $- g$ −g).
Greatest height: vertical velocity is 0, so $H = \frac{U ^{2} sin ^{2} α}{2 g}$ H = U²sin²α/2g. The speed there is $U cos α$ Ucos α, not 0.
Time of flight on level ground: $y = 0$ y = 0 again, so $T = \frac{2 U sin α}{g}$ T = (2Usin α)/g.
Range: $R = U cos α \times T = \frac{U ^{2} sin 2 α}{g}$ R = Ucos α × T = (U²sin 2α)/g, greatest when $sin 2 α = 1$ sin 2α = 1, so $α = 4 5^{\circ}$ α = 45°.

Worked example Mechanics · projectiles

A kicked ball (A2 only)

Question

A ball is kicked from level ground at 20 m/s at

3 0^{\circ}

above the horizontal. Find its time of flight, its range and its greatest height.

1
Components: horizontal $20 cos 3 0^{\circ} = 17.32$ 20cos 30° = 17.32, vertical $20 sin 3 0^{\circ} = 10$ 20sin 30° = 10
2
Time of flight: $0 = 10 t - 4.9 t^{2}$ 0 = 10t − 4.9t², so $t = \frac{10}{4.9} = 2.04$ t = 10/4.9 = 2.04
3
Range: $17.32 \times 2.041 = 35.3$ 17.32 × 2.041 = 35.3
4
Greatest height: $0 = 1 0^{2} - 2 (9.8) H$ 0 = 10² − 2(9.8)H, so $H = \frac{100}{19.6} = 5.10$ H = 100/19.6 = 5.10

Answer

2.04 s, 35.3 m, 5.10 m

⚠️ Watch out

multiplying the full 20 m/s by the time gives 40.8 m. Only the horizontal component carries the ball forwards.

Calculate

?(A2 only) A shot is released 2 m above level ground with speed 10 m/s at

4 0^{\circ}

above the horizontal. Find the horizontal distance it travels before hitting the ground. Give your answer to 3 significant figures.

metres

Mechanics · projectiles

The path and the model (A2 only)

Eliminate $t$ t using $t = \frac{x}{U c o s α}$ t = x/(Ucos α) to get the equation of the trajectory, a parabola:

y = x tan α - \frac{g x ^{2}}{2 U ^{2} cos ^{2} α} or y = x tan α - \frac{g x ^{2} ( 1 + tan ^{2} α )}{2 U ^{2}}

The model treats the ball as a particle, ignores air resistance and spin, and takes $g$ g as constant.
$4 5^{\circ}$ 45° is best only when the ball lands at its launch height. Launched from above the landing point, a slightly lower angle goes further.

Calculate

?(A2 only) A ball is projected from a point

O

on level ground with speed 20 m/s at an angle

α

above the horizontal, where

tan α = \frac{3}{4}

. Find the height of the ball when it has travelled 16 m horizontally. Give your answer as a decimal.

metres

Sort it

?In the standard projectile model (a particle, no air resistance), is each statement true or false?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?What angle throws a ball furthest, and why isn't the real answer that?

Recap

Big ideas

Velocity–time graph: gradient is acceleration, area is displacement.
For constant acceleration, list $s, u, v, a, t$ s, u, v, a, t and choose the suvat formula that leaves out the one you don't need.
Free fall: $a = g = 9.8$ a = g = 9.8 m s $^{- 2}$ ⁻² downwards; fix a positive direction and give every quantity its sign.
Down the chain differentiate: $s \to v \to a$ s → v → a. Up the chain integrate, and find each constant from the initial conditions.
At rest: $v = 0$ v = 0. Greatest or least velocity: $a = 0$ a = 0. Distance travelled: split the integral where $v$ v changes sign.
With vectors, work on each component; speed is $∣ v ∣$ |v|.
Split the motion: horizontal velocity $U cos α$ Ucos α stays constant; vertical motion has acceleration $g$ g downwards.
Time links the two: find $t$ t from the vertical motion, then use it horizontally.
Range $\frac{U ^{2} s i n 2 α}{g}$ (U²sin 2α)/g is greatest at $4 5^{\circ}$ 45° only on level ground in the no-air-resistance model.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Kinematics for CCEA GCE Mathematics (2210). 🎉

Mission 1A phone drops from a 45 m tower. When does it land, and how fast?
Mission 2A sprinter has $v = 3 t - 0.25 t^{2}$ v = 3t − 0.25t². When does she stop accelerating?
Mission 3What angle throws a ball furthest, and why isn't the real answer that?

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Specification points and sources for this lesson are listed at the bottom of the page.

Kinematics

A phone drops from a 45 m tower. When does it land, and how fast?

The language and the graphs

Reading a velocity–time graph

The constant acceleration formulae

Stopping distance

Vertical motion under gravity

A ball thrown up from a cliff

A sprinter has v=3t−0.25t2v = 3t − 0.25t². When does she stop accelerating?

Differentiate to go from displacement to acceleration

When is the particle at rest?

Integrate to go back up

From acceleration to displacement

Calculus with vectors (A2 only)

What angle throws a ball furthest, and why isn't the real answer that?

Constant acceleration with vectors

Thrown horizontally from a cliff

Time of flight, greatest height and range (A2 only)

A kicked ball (A2 only)

The path and the model (A2 only)

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

A sprinter has $v = 3 t - 0.25 t^{2}$ v = 3t − 0.25t². When does she stop accelerating?