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: 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`