Draw a velocity time graph for an uniformly accelerated object using velocity time graph drive V =u+at
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Derive v = u + at by Graphical MethodConsider the velocity – time graph of a body shown in the below Figure.Velocity–Time graph to derive the equations of motion.The body has an initial velocity u at point Aand then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.Now, Initial velocity of the body, u=OA...... (1)And, Final velocity of the body, v=BC........ (2)But from the graph BC=BD + DCTherefore, v=BD + DC ......... (3)Again DC=OASo, v=BD + OANow, From equation (1), OA=uSo, v=BD + u ........... (4)We should find out the value of BD now. We know that the slope of a velocity – time graph is equal to acceleration, a.Thus, Acceleration, a=slope of line ABor a=BD/ADBut AD=OC = t,so putting t in place of AD in the above relation, we get:a=BD/tor BD=atNow, putting this value of BD in equation (4) we get :v=at + uThis equation can be rearranged to give:v=u + at
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the variation in velocity with time for an object moving in a straight line can be represented by a velocity time graph in this graph time is represented along the x axis and the velocity is represented along y axis if the object moves to uniform velocity the height of the graph will not change it will be the straight line parallel to x axis so the distance s moved by the car in time [t2 -t1]
s=AC *AD [*MEANS INTO]
=[(40KM PER HR) * [T2-T1)H]
=40 (T2-T1)KM THIS IS YOUR ANSWER
s=AC *AD [*MEANS INTO]
=[(40KM PER HR) * [T2-T1)H]
=40 (T2-T1)KM THIS IS YOUR ANSWER
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