The graph in ?Fig. E2.31? shows the velocity of a motorcycle police officer plotted as a function of time. (a) Find the instantaneous acceleration at t = 3 s, t = 7 s, and t = 11 s. (b) How far does the officer go in the first 5 s? The first 9 s? The first 13 s?

Solution 31E Step 1 of 5: (a) ind the instantaneous acceleration at t = 3 s, t = 7 s, and t = 11 s. To calculate the instantaneous acceleration at the given time, we need to calculate the slope of the tangent at the given time on the v-t curve. Instantaneous acceleration at t = 3s, From the given graph, if we see the curve at t= 3s; the velocity is constant. Also the slope of the line is zero(tangent at that point is horizontal), which means the acceleration is zero. At t =3s, a= 0 Therefore, the instantaneous acceleration at t= 3s is zero. Instantaneous acceleration at t = 7s, Tangent at t= 7s is drawn as shown below, Slope of the tangent gives instantaneous acceleration, At t=7s, v=30m/s and at t =6s , v= 25 m/s Therefore 30 m/s 25 m/s Slope=acceleration= 7s 6s 2 a is5 m/s Therefore, the instantaneous acceleration at t= 7s is 5 m/s . 2