the displacement (in feet) of a certain particle moving in a straight line is gi

Question

the displacement (in feet) of a certain particle moving in a straight line is given by S=t^3/8 where t is measured in seconds. find the average velocity over the interval [1,1.17]

Explanation / Answer

s = (t^3)/8 velocity is the differentiation of displacement in time.. velocity = ds/dt so differentiating both sides by 't' we get ds/dt = 3*t^2/8 (which is the velocity) substitutin time interval we get for t = 1.8 => velocity v = 3*1.8^2/8 = 1.215 for t = 1 => initial vel. v = 3*1/8 = 0.375 avg velocity = 0.795 It is important to see here that v(t) is a continuous function, and you must use calculus to find the average of a continuous function. the average of f(x) over an interval [a,b] is: Integral[f(x) dx]/Integral[dx] with the limits a, b this can be written: Integral[f(x) dx]/(b-a) Here, we know that s(t)=t^3/8, so v(t)=s'(t) =3t^2/8 now, the average value of v(t) over the interval is: Integral[3t^2/8 dt]/(1.8-1) =t^3/8/0.8 with t evaluated at 1.8 and 1 (not surprisingly, when we integrate v(t) we get back s(t)) so the average value is: 1/8[1.8^3-1^3]/0.8=0.755 ft/s

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