Question

derive 3 equation of motion writh the help of a diagram

Answer 1

Consider the velocity-time graph of an object that moves under uniform acceleration as shown in Fig From this graph, you can see thatinitial velocity of the object is u (at point A)and then it increases to v (at point B) in timet. The velocity changes at a uniform rate a. the perpendicular lines BC andBE are drawn from point B on the time andthe velocity axes respectively, so that theinitial velocity is represented by OA, the finalvelocity is represented by BC and the timeinterval t is represented by OC. BD = BC –CD, represents the change in velocity in timeinterval t.Let us draw AD parallel to OC. From thegraph, we observe thatBC = BD + DC = BD + OASubstitutingBC=v and OA = u,we getv=BD + uorBD=v – u
From the velocity-time graph (Fig. 8.8),the acceleration of the object is given bya = Changeinvelocitytimetaken=BDBD=ADOCSubstituting OC = t, we geta = BDtorBD = at(8.9)Using Eqs. (8.8) and (8.9) we getv = u + at8.5.2EQUATION FOR POSITION-TIMERELATIONLet us consider that the object has travelleda distance s in time t under uniformacceleration a. In Fig. 8.8, the distancetravelled by the object is obtained by the areaenclosed within OABC under the velocity-timegraph AB.Thus, the distance s travelled by the objectis given bys= area OABC (which is a trapezium)= area of the rectangle OADC + area ofthe triangle ABD
= OA × OC + 12 (AD × BD)(8.10)Substituting OA = u, OC = AD = t andBD = at, we gets = u × t + 1()2t×ators = u t + 12 a t 28.5.3EQUATION FOR POSITION–VELOCITYRELATIONFrom the velocity-time graph shown inFig. 8.8, the distance s travelled by the objectin time t, moving under uniform accelerationa is given by the area enclosed within thetrapezium OABC under the graph. That is,s = area of the trapezium OABC= ()OA+BC×OC2Substituting OA = u, BC = v and OC = t,we get()2=u+vts(8.11)From the velocity-time relation (Eq. 8.6),we gett=vua()(8.12)Using Eqs. (8.11) and (8.12) we have()()×v+uv-us=2aor 2 a s = v2 – u2



or you can refer to page no 107 and 108 in science ncert.