Calculus for Kinematics | DP IB Maths: AI HL Revision Notes 2021

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Differentiation for Kinematics

Displacement, velocity and acceleration are related by calculus
In terms of differentiation and derivatives
- velocity is the rate of change of displacement
  - $v = \frac{d s}{d t}$ or $v (t) = s' (t)$
- acceleration is the rate of change of velocity
  - $a = \frac{d v}{d t}$ or $a (t) = v' (t)$
- so acceleration is also the second derivative of displacement
  - $a = \frac{d^{2} s}{d t^{2}}$ or $a (t) = s'' (t)$
- Sometimes velocity may be a function of displacement rather than time
  - $v (s)$ rather than $v (t)$
  - in such circumstances, acceleration is $a = v \frac{d v}{d s}$
  - this result is derived from the chain rule
- All acceleration formulae are given in the formula booklet
Even if a motion graph is given, if possible, use your GDC to draw one
- you can then use your GDC’s graphing features to find gradients
  - velocity is the gradient on a displacement (-time) graph
  - acceleration is the gradient on a velocity (-time) graph
Dot notation is often used to indicate time derivatives
- $x$ is sometimes used as displacement (rather than $s$ ) in such circumstances
- $\dot{x} = \frac{d x}{d t}$ , so $\dot{x}$ is velocity
- $\overset{"}{x} = \frac{d^{2} x}{d t^{2}}$ , so $\overset{"}{x}$ is acceleration

The displacement,

x

m, of a particle at

t

seconds, is modelled by the function

x (t) = 2 t^{3} - 27 t^{2} + 84 t

Find expressions for

\dot{x}

and

\overset{"}{x}

The velocity,

v

m s^-1, of a particle is given as

v (s) = 6 s - 5 s^{2} - 4

, where

s

m is the displacement of the particle.

Find an expression, in terms of

s

, for the acceleration of the particle.

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Since velocity is the derivative of displacement ( $v = \frac{d s}{d t}$ ) it follows that

$s = \int v d t$

$v = \int a d t$

You might be given the acceleration in terms of the velocity and/or the displacement
- In this case you can solve a differential equation to find an expression for the velocity in terms of the displacement

$a = v \frac{d v}{d s}$

A boundary or initial condition would need to be known
- phrases involving the word “initial”, or “initially” are referring to time being zero, i.e. $t = 0$
- you might also be given information about the object at some other time (this is called a boundary condition)
- substituting the values in from the initial or boundary condition would allow the constant of integration to be found

Two graphs showing the difference between the graph (and integral) of v=v(t) and the graph (and integral) of v=|v(t)|

Sketching the velocity-time graph can help you visualise the distances travelled using areas between the graph and the horizontal axis

A particle moving in a straight horizontal line has velocity ( $v m s^{- 2}$ ) at time $t$ seconds modelled by $v (t) = 8 t^{3} - 12 t^{2} - 2 t$ .

Given that the initial position of the particle is at the origin, find an expression for its displacement from the origin at time $t$ seconds.
Find the displacement of the particle from the origin in the first five seconds of its motion.
Find the distance travelled by the particle in the first five seconds of its motion.

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