Sec 4 · Additional Mathematics

Kinematics

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1. Displacement, velocity and acceleration

Find	Differentiate	Integrate
velocity $v$	$v = \dfrac{ds}{dt}$	$v = \int a \, dt$
acceleration $a$	$a = \dfrac{dv}{dt}$
displacement $s$		$s = \int v \, dt$

Words	Means
at $O$	$s = 0$
instantaneously at rest	$v = 0$
initial	$t = 0$
maximum or minimum velocity	$a = 0$

At rest: s = t³ − 12t² + 45t

\displaystyle v = 3t^{2} - 24t + 45

e.g.

\displaystyle 3t^{2} - 24t + 45 = 0

\displaystyle t^{2} - 8t + 15 = 0

\displaystyle (t - 3)(t - 5) = 0

\displaystyle t = 3 \text{ or } t = 5

📖 Worked example

A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$ , its displacement from $O$ , $s$ m, is given by $s = t^{3} - 12t^{2} + 45t$ . Find the values of $t$ at which the particle is instantaneously at rest.

Step 1 of 4

Find v from s: differentiate or integrate?

v = \dfrac{ds}{dt}

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2. Integrating to find velocity and displacement

v = 6e⁻²ᵗ, starting at O

\displaystyle s = -3e^{-2t} + c

e.g.

\displaystyle 0 = -3e^{0} + c

\displaystyle c = 3

$s = 3 - 3e^{-2t}$

📖 Worked example

A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$ , its velocity, $v$ m/s, is given by $v = 6e^{-2t}$ . Find its exact displacement from $O$ when $t = 1$ .

Step 1 of 3

Displacement: s = ∫ v dt (e to the power kt integrates to itself divided by k)

s = -3e^{-2t} + c

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3. Distance travelled

v = t² − 7t + 6 in the first 5 seconds: it turns back at t = 1

\displaystyle s = \dfrac{1}{3}t^{3} - \dfrac{7}{2}t^{2} + 6t

Velocity-time graph of v = t² − 7t + 6 from t = 0 to 5. The area above the axis up to t = 1 is 17/6; below the axis from t = 1 to 5 the area is 56/3, when the particle moves back.

$t = 0$ to $1$ : $\dfrac{17}{6}$ m
$t = 1$ to $5$ : $\dfrac{56}{3}$ m (moving back)
distance: $\dfrac{17}{6} + \dfrac{56}{3}$ , which is $\dfrac{43}{2}$ m

📖 Worked example

A particle moves in a straight line so that, $t$ seconds after passing through a fixed point $O$ , its velocity, $v$ m/s, is given by $v = t^{2} - 7t + 6$ . Find the total distance travelled by the particle in the first $5$ seconds.

Step 1 of 4

Does it turn back? Find when v = 0

t^{2} - 7t + 6 = 0

(t - 1)(t - 6) = 0

t = 1

t = 6

Practise this →

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Practice questions by skill

1. Displacement, velocity and acceleration

Find	Differentiate	Integrate
velocity $v$	$v = \dfrac{ds}{dt}$	$v = \int a \, dt$
acceleration $a$	$a = \dfrac{dv}{dt}$
displacement $s$		$s = \int v \, dt$

Words	Means
at $O$	$s = 0$
instantaneously at rest	$v = 0$
initial	$t = 0$
maximum or minimum velocity	$a = 0$

At rest: s = t³ − 12t² + 45t

\displaystyle v = 3t^{2} - 24t + 45

e.g.

\displaystyle 3t^{2} - 24t + 45 = 0

\displaystyle t^{2} - 8t + 15 = 0

\displaystyle (t - 3)(t - 5) = 0

\displaystyle t = 3 \text{ or } t = 5

📖 Worked example

Step 1 of 4

Find v from s: differentiate or integrate?

v = \dfrac{ds}{dt}

Practise this →

2. Integrating to find velocity and displacement

v = 6e⁻²ᵗ, starting at O

\displaystyle s = -3e^{-2t} + c

e.g.

\displaystyle 0 = -3e^{0} + c

\displaystyle c = 3

$s = 3 - 3e^{-2t}$

📖 Worked example

Step 1 of 3

Displacement: s = ∫ v dt (e to the power kt integrates to itself divided by k)

s = -3e^{-2t} + c

Practise this →

3. Distance travelled

v = t² − 7t + 6 in the first 5 seconds: it turns back at t = 1

\displaystyle s = \dfrac{1}{3}t^{3} - \dfrac{7}{2}t^{2} + 6t

$t = 0$ to $1$ : $\dfrac{17}{6}$ m
$t = 1$ to $5$ : $\dfrac{56}{3}$ m (moving back)
distance: $\dfrac{17}{6} + \dfrac{56}{3}$ , which is $\dfrac{43}{2}$ m

📖 Worked example

Step 1 of 4

Does it turn back? Find when v = 0

t^{2} - 7t + 6 = 0

(t - 1)(t - 6) = 0

t = 1

t = 6

Practise this →