Differentiation and Integration Relating Position, Velocity, and Acceleration (1D)

Instantaneous vs. Average Rate of Change

• When referring to Rate of Change (ROC) we can refer to both Instantaneous ROC and Average ROC
• Instantaneous ROC involves taking the derivative of a function and finding the slope at that specific point
• AKA the Tangent Line
• Average ROC involves finding the slope between two points on a function
• AKA the Secant Line

• Where x(t) is the displacement function of a particle or object with respect to time Example: Find the instantaneous velocity at t = 2s $x\left(t\right)\ =\ x^3+x^2+x$ $\frac{dx}{dt}=\ v\left(t\right)\ =\ 3x^2+2x+1$ $v\left(2\right)\ =\ 3\left(2\right)^2+2\left(2\right)+1$ $v\left(2\right)\ =\ 17\left(\frac{m}{s}\right)$

• Where x(t) is the displacement function of a particle or object with respect to time Example: Find the average velocity from t = 1s to t = 3s for the following position function: $x\left(t\right)\ =\ x^3+x^2+x$ x(3) = 33+32+3 x(3) = 39 x(1) = 13+12+1 x(1) = 3 $v_{avg}=\frac{\left(x\left(3\right)-x\left(1\right)\right)}{3-1}$ $v_{avg}=\frac{\left(39-3\right)}{3-1}$ $v_{avg}=12\ \left(\frac{m}{s}\right)$

• $$Where v(t) is the velocity function of a particle or object with respect to time Example: Find the instantaneous acceleration at t = 2 using the following velocity function: $v\left(t\right)\ =\ 3x^2+2x$ $\frac{dv}{dt}=a\left(t\right)\ =\ 6x+2$$v\left(1\right)\ =\ 5$ $a\left(2\right)\ =\ 6\left(2\right)\ +1\$ $a\left(2\right)\ =\ 13$

• Where v(t) is the velocity function of a particle or object with respect to time Example: Find the average acceleration from t = 1 to t = 3 for the following velocity function: $v\left(t\right)\ =\ 3x^2+2x$ $v\left(3\right)\ =\ 3\left(3\right)^2+2\left(3\right)$ $v\left(3\right)\ =\ 33$ $v\left(1\right)\ =\ 3\left(1\right)^2+2\left(1\right)$ $v\left(1\right)\ =\ 5$ $a_{avg}=\frac{\left(v\left(3\right)-v\left(1\right)\right)}{3-1}$ $a_{avg}=\frac{\left(33-5\right)}{3-1}$ $a_{avg}=14\ \left(\frac{m}{s^2}\right)$