In example 1, the method we find the acceleration function is called find the derivative of the velocity function. In Calculus there is a simple way to find the derivative of a function. Give Functions where `C` is a constant. The derivative of this function is: Notes: the constant `C` just dropped. So in example 1 the function of velocity is `v_x = 40 - 5t^2` `a_x = (dv_x)/(dt) = (-5)(2)t^(2-1) = -10t` More practice: `f(x) = 3x^3 + 4x^2 - x +5` `f'(x) = (3)(3)x^(3-1) + (4)(2)x^(2-1) +(-1)(1)x^(1-1)` `f'(x) = 9x^2 + 8x -1` `x = 100 + 20t + 5t^2` `(dx)/(dt) = 20 +(5)(2)t = 20+ 10t` Example 2: The displacement of a particle moving along the `x` axis varies according to the expression `x = 10 - 5t^2+4t^3`, where `x` is in meters and `t` is in seconds. Find displacement, velocity and acceleration at `t=2.0\ s` Displacement at `t=2.0\ s`: `x = 10 - 5(2.0)^2 + 4(2.0)^3 = 10-20 + 32 = 22\ m` Velocity is the derivative of displacement function: `v = (dx)/(dt) = (-5)(2)t + (4)(3)t^2 = -10t + 12t^2` Velocity at `t = 2.0\ s` `v = (-10)(2.0) + 12(2.0)^2 = -20 + 48 = 24\ m//s` Acceleration is the derivative of velocity function: `a = (dv)/(dt) = -10 + 24t ` Acceleration at `t = 2.0\ s` `a = -10 + 24(2.0) = 38\ m//s^2`