# relationship between kinematics and area under curve

My teacher is giving us an honors pre-calc/calculus introduction to physics and the kinematics learning target he has given us is to understand the "relationship between area under a curve and the equation that calculates motion". I understand that the area under a curve (from the curve to the x-axis) will calculate the slope of the curve (which makes absolutely no mathematical sense to me), but I don't know how that relates to calculating motion.

Please understand that my teacher is a very strict, high-standards, superfluous teacher who works at a university speed (he says so himself, and I'm only in HS) so there may be gaps in my understanding. I know what $\Delta$ means and I have a very, very loose understanding of the relationships between acceleration, velocity, and distance.

• Displacement NOT distance is the basic variable here. Displacement is a vector where distance is a scalar, i.e. displacement = distance & direction, e.g. 50m to the north.

• Velocity: Rate of change (with respect to time) of displacement. This means, that the slope of the displacement v/s time curve gives you the velocity at that instant in time.

• Acceleration: Rate of change of velocity. Slope of the velocity v/s time curve gives the acceleration at that instant in time.

Now, the rate of change is the derivative of that function. So,

$$\text{Displacement}=s;\\ \text{Velocity (v)}=\frac{ds}{dt};\\ \text{Acceleration (a)}=\frac{dv}{dt}=\frac{d}{dt}(\frac{ds}{dt})=\frac{d^2s}{ds^2};$$

Don't be scared by the $\displaystyle \frac{d^2s}{dt^2}$, that's just the way to denote the second derivative of something.

On the other hand, the inverse of the derivative is called the integral. Geometrically, integration of a function gives you the area under the graph of that function.

$$\text{Acceleration} = a;\\ \text{Velocity} = \int{a\ dt};\\ \text{Displacement} = \int{v\ dt} = \iint{a\ dt};$$

• I just started Pre-Calculus, so I have no idea about any Calculus concepts. (The course catalog lists it as friendly to people co-enrolled in Pre-Calc, but I guess that's more of a personal plight). Some of those concepts I know of include integration, integrals, derivatives, and that curious apostrophe that means prime. I know some of the names, but not what they are or how they work and I don't have the time to learn an entire year of math. On the other hand, I am not required to use that; I just 'count the squares' of the grid above and below the function and find it that way. – person27 Oct 18 '13 at 6:35
• I think adding some units will help the buddy. – user29727 Oct 18 '13 at 9:55
• Acceleration: $m \cdot s^{-2}$ Velocity: $m \cdot s^{-1}$ Displacement: $m$ The other way around: Displacement: $m$ Velocity: $m \cdot s^{-1}$ Acceleration: $m \cdot s^{-2}$ Easy as drinking water. – user29727 Oct 18 '13 at 9:57

You should explain what part you understand.

The slope is the derivate, so for y(x), y'(x) indicates how fast the value y(x) is changing. Velocity is the derivate of position wrt time and acceleration is the derivate of velocity.

The area under the curve of y(x) gives you the "opposite" of the slope. It is called the integral of y respect to x. For example, if y=velocity and x=t, the area would give you the distance travelled.