## Instantaneous Velocity and Speed (ADVANCED)

The

**instantaneous velocity**`v_x` equals the limiting value of the ratio `(∆x)/(∆t)` as `∆t` approaches zero: `v_x = lim_(∆t->0)(∆x)/(∆t)=(dx)/(dt)` In calculus notation, this limit is called the derivative of `x` with respect to `t`, written `(dx)/(dt)`: The**instantaneous speed**of a particle is defined as the magnitude of its instantaneous velocity. Example: A particle moves along the `x` axis. Its position varies with time according to the expression `x = -4t + 2t^2`, where `x` is in meters and `t` is in seconds. The position–time graph for this motion is shown in above Figure. (A) Determine the displacement of the particle in the time intervals `t = 0` to `t = 1 s` and `t = 1 s` to `t = 3 s`. Solution: Given `x = -4t + 2t^2` 1. in the time intervals `t = 0` to `t = 1 s` (`A->B`) When `t=0` we get `x_i = -4(0) + 2(0)^2 = 0` When `t=1` we get `x_f = -4(1) + 2(1)^2 = -2 m` `∆x_(A->B) = x_f - x_i = -2 - 0 = -2 m` 2. in the time intervals `t = 1` to `t = 3 s`(`B->D`) When `t=1` we get `x_i = -4(1) + 2(1)^2 = -2 m` When `t=3` we get `x_f = -4(3) + 2(3)^2 = 6 m` `∆x_(B->C) = x_f - x_i = 6 - (-2) = 8 m` (B) Calculate the average velocity during these two time intervals. Solution: 1. in the time intervals `t = 0` to `t = 1 s` (`A->B`) `∆t = t_f-t_i = 1-0 = 1s` `v_(avg)=∆x_(A->B)/∆t = (-2 m) / (1 s) = -2 m//s` which is same as the slope of line `AB` 2. in the time intervals `t = 1` to `t = 3 s`(`B->D`) `∆t = t_f-t_i = 3-1 = 2 s` `v_(avg)=∆x_(B->D)/∆t = (8 m) / (2 s) = 4 m//s` which is same as the slope of line `BD` (C) Find the instantaneous velocity of the particle at `t = 2.5 s`. Solution 1: by measure The instantaneous velocity at point `C` is the slop of tangent line at point `C`. In above Figure, the green line is the tangent line at `t = 2.5 s` Measure the slope of the green line we get `v_(avg) = (10m - (-4)m)/(3.8s-1.5s) = 6 m//s` Solution 2: by calculus Given `x = -4t + 2t^2` `v_x = (dx)/(dt) = -4 + 2(2)t= -4 +4t` at `t=2.5s` `v_x = -4 + (4)(2.5) = 6 m//s`