Derive equation for position-time relation
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Let the initial velocity of the object = u Let the object is moving with uniform acceleration, a. Let object reaches at point B after time, t and its final velocity becomes, vDraw a line parallel to x-axis DA from point, D from where object starts moving. Draw another line BA from point B parallel to y-axis which meets at E at y-axis.
Let OE = time, t Now, from the graph, BE = AB + AE ⇒ v = DC + OD (Since, AB = DC and AE = OD) ⇒ v = DC + u (Since, OD = u) ⇒ v = DC + u ------------------- (i) Now, Acceleration (a) =(Change in velocity)/(Time taken) a= v – ut⇒ a = v - ut ⇒ a= OC – OD t = DC t ⇒a = OC-OD t =DC t ⇒at=DC ⇒at=DC -----(ii) By substituting the value of DC from (ii) in (i) we get v= at + u v = at + u ⇒ v = u + at ⇒ v = u + at Above equation is the relation among initial velocity (u), final velocity (v), acceleration (a) and time (t). It is called first equation of motion.
Let OE = time, t Now, from the graph, BE = AB + AE ⇒ v = DC + OD (Since, AB = DC and AE = OD) ⇒ v = DC + u (Since, OD = u) ⇒ v = DC + u ------------------- (i) Now, Acceleration (a) =(Change in velocity)/(Time taken) a= v – ut⇒ a = v - ut ⇒ a= OC – OD t = DC t ⇒a = OC-OD t =DC t ⇒at=DC ⇒at=DC -----(ii) By substituting the value of DC from (ii) in (i) we get v= at + u v = at + u ⇒ v = u + at ⇒ v = u + at Above equation is the relation among initial velocity (u), final velocity (v), acceleration (a) and time (t). It is called first equation of motion.
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