## Acceleration

An particle whose velocity is changing is said to be accelerating. For instance, a car whose velocity increases from 0 to 50\ km//h is accelerating. The average acceleration a_(x,avg) of the particle is defined as the change in velocity ∆v_x divided by the time interval ∆t during which that change occurs:
a_(x,avg) = (∆v_x)/(∆t) = (v_(xf) - v_(x i))/(t_f - t_i)
Graphical Relationships Between x, v_x, and a_x Average and Instantaneous Acceleration: Example 1: The velocity of a particle moving along the x axis varies according to the expression v_x = 40 - 5t^2, where v_x is in meters per second and t is in seconds. (A) Find the average acceleration in the time interval t = 0 to t = 2.0 s. Solution 1. at t=0 V_A = 40 - 5(0)^2 = 40 m//s  2. at t=2 V_B = 40 - 5(2)^2 = 20 m//s  3. a_avg = (v_B - v_A)/(∆t) = ( 20 - 40)/(2-0) = -10 m//s^2 (B) Determine the acceleration at t = 2.0 s. Solution 1. Knowing that the initial velocity at any time t is v_(x i) = 40 - 5t^2, find the velocity at any later time t + ∆t: v_(xf) = 40- 5(t+∆t)^2  = 40- 5(t^2 + 2t∆t + (∆t)^2)  = 40- 5t^2 - 10t∆t - 5(∆t)^2  = v_(x i) - 10t∆t - 5(∆t)^2 ∆v = v_(xf)-v_(x i)   = -10t∆t - 5(∆t)^2 To find the acceleration at any time t, divide this expression by ∆t and take the limit of the result as ∆t approaches zero: a_x = lim_(∆t->0)(∆V_x)/(∆t) =  = lim_(∆t->0)(-10t∆t - 5(∆t)^2)/(∆t) = lim_(∆t->0)(-10t - 5∆t) =-10t Substitute t = 2.0 s: a_x = (-10)(2.0) = -20 m//s^2