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 Step 2 of 5: Instantaneous acceleration at t = 11s, Similarly Tangent at t= 11s is drawn as shown below, Slope of the tangent gives instantaneous acceleration, At t=11s, v=25m/s and at t =12s , v= 10 m/s Therefore 10 m/s 25 m/s Slope=acceleration= 12 s 11 s 2 ais -15 m/s Therefore, the instantaneous acceleration at t= 11s is -15 m/s . 2 Step 3 of 5: (b) How far does the officer go in the first 5 s The first 9 s The first 13 s We know that, the area under the curve in the velocity -time graph will given the displacement. Displacement of cat in first 5s. The area under the curve until 5 seconds is, The area of the rectangle, length=5 s and height=20 m/s Area = length×height = 5 s ×20 m/s = 100 m Therefore, the officer travels 100 m in first 5 seconds.