Do you recollect studying calculus in high school? Does the concept of differentiation and integration sound familiar? When we were in school, ‘Derivatives’ held a different meaning – it was just the process of working through challenging integration and differentiation equations.
Let me try to jog your memory – the aim here is to get the point across without getting too deep into the nuts and bolts of solving a calculus issue. It’s essential to keep on reading, as this discussion is specifically related to options.
Consider the following scenario:
– Let ‘x’ represent the distance and ‘dx’ represent the change in distance. The change in distance, or ‘dx’, is calculated as 3 (8 – 5).
– Similarly, let ‘t’ represent time and ‘dt’ represent the change in time. The change in time, or ‘dt’, is 8 (20 – 12).
– Dividing the change in distance by the change in time gives us the velocity, denoted as ‘V’.
– V = dx / dt = 3/8.
– The velocity is expressed in kilometres per minute.
– V = 3 / (8/60) = (3 * 60) / 8 = 22.5 KMPH.
– Hence, the car is moving at a velocity of 22.5 KMPH (kilometres per hour).
– The speed of the car during the first part of the journey was 22.5 KMPH. By calculating the change in distance (dx=6) and the change in time (dt=6) during the second part, we can determine its velocity.
– The velocity for the second part is calculated as 6 KMPH (dx=6 and dt=6).
– The change in velocity can be determined by subtracting the initial velocity (22.5 KMPH) from the final velocity (6 KMPH).
– The change in velocity is 6 KMPH – 22.5 KMPH = -16.5 KMPH.
– The negative sign indicates a decrease in velocity, suggesting deceleration.
We made things much easier by assuming that acceleration is constant. Of course, in reality, this isn’t the case as you naturally accelerate at different speeds. If such a problem involves changes in one variable due to another, one must delve into a branch of calculus which is known as Differential Equations.
Let’s consider the following now:
Distance travelled (position) changes according to velocity, which is referred to as the 1st order derivative of position.
Change in Velocity = Acceleration
Acceleration is the rate at which velocity changes, which is itself the rate of change in position.
Therefore, Acceleration can be said to be the rate of change of Velocity or the second derivative of Position.
Bear in mind the 1st and 2nd derivatives as we continue to analyse Gamma.