International Baccalaureate Mathematics

Acceleration	$ a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2} $
Distance travelled from $t_1 \text{ to } t_2$	$ \int^{t_2}_{t_1}\|v(t)\|dt$
Displacement from $t_1 \text{ to } t_2$	$ \int^{t_2}_{t_1}v(t)dt$

Kinematic problems in the realm of calculus involve the study of motion of objects using differentiation and integration. These mathematical tools allow us to derive relationships between displacement $ s $, velocity $ v $, acceleration $ a $, and the total distance travelled. Displacement represents the overall change in position of an object, velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity with respect to time. Conversely, the total distance travelled can be found by integrating the velocity function over a given time interval.

The key formulae for kinematics in calculus are:

$$ v(t) = \frac{ds}{dt} \\ a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} \\ s_{\text{total}} = \int_{t_1}^{t_2} v(t) \, dt $$

Where:

$ v(t) $ is the velocity as a function of time,
$ a(t) $ is the acceleration as a function of time,
$ s $ is the displacement, and
$ s_{\text{total}} $ is the total distance travelled.

For example, consider an object with a velocity function given by $ v(t) = 2t $. The acceleration, being the derivative of velocity, is:

$$ a(t) = \frac{dv}{dt} = 2 \text{ m/s}^2 $$

To find the total distance travelled by the object from $ t = 1 $ second to $ t = 3 $ seconds, we integrate the velocity function:

$$ s_{\text{total}} = \int_{1}^{3} 2t \, dt = [t^2]_1^3 = 9 - 1 = 8 \text{ metres} $$

Thus, the object has travelled a total distance of 8 metres between 1 and 3 seconds.

This video on Kinematics is from Revision Village and is aimed at students taking the IB Maths AA SL/HL course

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Acceleration	\( a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2} \)
Distance travelled from \(t_1 \text{ to } t_2\)	\( \int^{t_2}_{t_1}\|v(t)\|dt\)
Displacement from \(t_1 \text{ to } t_2\)	\( \int^{t_2}_{t_1}v(t)dt\)