. .1...\" -...- -u---..- .u. ---.-..- \"Unit 1 KINEMATICS Areas UNDER Velocity Graphs Suppose we hash to line the age between the graph and the time axis on the graph below between t = tn and t: t1. The area = height 1 base =[viift1- to} = 1h * M v [mils] But from the denition of 1new: a: = V - at. if tr is eenstant over the interval at. Therefore we can see that the area between the v vs. t graph and. the t axis gives the disglaeer'nent of the body during the time interval L Dr in short: Area (vvs.ti=ex = itrI=V1I[tr 49] Area {tr vs. ti = the displacement between t = t, and t = t1. Referring to the graph above, answer the following questions: 1. 1Il'tfhat does the area under the graph and the t axis between t1 and t2 represent?I 2. I|.|'||.I'hat is the displacement of the object between t; and t3? 11123 = 3. What is the displacement at the object between tn and h? le4: UHII | ll'llzl'v'll [Lab 4. Suppose the graph on the previous page is described by vft} = ltt2 + in between t4 and t5. How would you nd the displacement between t4 and t5? Since velocity is the derivative of the position with respect to time, through calculus, we can also nd the displacement by nding a position function whose derivative ts the function for velocity. This process is called integration {or andifferentiation]. 1: | DISPLACEMENT from a function ofvelocity is given by ex = {x2 X1}"'I The above is just a fancy way of calcuiating the AREA BETWEEN THE VELOCITY VERSUS TIME AXIS BETWEEN TIMES 1'1 and t2. So to nd the displacement between L; and t5, you must integrate the above velocity function with limits of integration between L; and is. ex\" = J vtt) dt = I\" cit = 5. a} From the graph on the previous page what is the displacement between t5 and t6? b) Your answer in part a} should be negative. 1vlv'hat physical meaning does this negative sign mean for the object? E. How would you nd the total displacement of the object between to and iv