Calculus for Kinematics (DP IB Analysis & Approaches (AA): HL): Revision Note

Differentiation for kinematics

How is differentiation used in kinematics?

• Displacement, velocity and acceleration are related by calculus

• In terms of differentiation and derivatives

  • velocity is the rate of change of displacement

    • $v=\frac{\mathrm{d}s}{\mathrm{d}t}$  or  $v\left(t\right)=s\text{'}\left(t\right)$

  • acceleration is the rate of change of velocity

    • $a=\frac{\mathrm{d}v}{\mathrm{d}t}$  or  $a\left(t\right)=v\text{'}\left(t\right)$

  • so acceleration is also the second derivative of displacement

    • $a=\frac{{\mathrm{d}}^{2}s}{\mathrm{d}{t}^2}$  or  $a\left(t\right)=s\text{'}\text{'}\left(t\right)$

• If a graph is not given you can 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

Integration for kinematics

 How is integration used in kinematics?

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

$s=\int v\mathrm{d}t$

• Similarly, velocity will be an antiderivative of acceleration

$v=\int a\mathrm{d}t$

How would I find the constant of integration in kinematics problems?

• 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 allows the constant of integration to be found

How are definite integrals used in kinematics?

• Definite integrals can be used to find the displacement of a particle between two points in time

  • ${\int }_{t_1}^{t_2}v\left(t\right)\mathrm{d}t$ would give the displacement of the particle between the times$t=t_1$ and$t=t_2$

    • This can be found using a velocity-time graph by subtracting the total area below the horizontal axis from the total area above

  • ${\int }_{t_1}^{t_2}\left|v\left(t\right)\right|\mathrm{d}t$ gives the distance a particle has travelled between the times$t=t_1$ and$t=t_2$

    • This can be found using a velocity velocity-time graph by adding the total area below the horizontal axis to the total area above

    • You can use a GDC to plot the modulus graph$y=\left|v\left(t\right)\right|$