Unit 1 KINEMATICS 3. What is the change in velocity of the object between to and t? AV0-6= 4. Suppose the graph on the previous page is described by a(t) = kt + b How would you find the change in velocity between t6 and t7? between to and t7. Since acceleration is the derivative of the velocity with respect to time, through calculus, we can also find the velocity by finding a velocity function whose derivative fits the function for acceleration. Again, we use process of integration. Change in Velocity from a function of acceleration is given by Av = (V- V) = a(t)dt where the resulting sign indicates the direction of the CHANGE in the velocity vector. The above is just a fancy way of calculating the AREA BETWEEN THE ACCELERATION VERSUS TIME AXIS BETWEEN TIMES t and t. 5. So to find the change in velocity between t and t7, you must integrate the above velocity function with limits of integration between to and t7. Av6-7 = a(t) dt = AV 6-7= dt = Areas UNDER Acceleration Graphs Given the acceleration versus time graph below, what is the area between the acceleration line and the time axis between t = to and t = t? Area from to to t =. a (m/s)4 a1 11 S 14 15 t(s) t7 a2 Your answer should have been Area from to to t = a (tto) a At But from the definition of a ave: Av=a. At, if a is constant over the interval At. Therefore, we can see that the area between the a vs. t graph and the t axis gives the change in velocity of the body during the time interval to to t. Or in short: Area (a vs. t) = Av = V - Vo = a (t - to) Area (a vs. t) = the change in velocity between t = to and t = t. Referring to the graph above, answer the following questions: Look familiar? 1. What does the area under the graph and the t axis between t and t represent? 2. What is the change in velocity of the object between t and t3? AV2-3 =