## 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