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The position s of a point (in feet) is given as a function of time t (in seconds). s = 2/t − 2/t^2; t = 2 (a) Find its acceleration as a function of t. s''(t) = (b) Find the point's acceleration at the specified time. s''(2) =
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$$a(x) = \frac{d^2 s}{dt^2}$$
Its s= \[\frac{ 2 }{ t }-\frac{ 2 }{ t^2 }\] t=2
s''= \[(4/t^{3})-(12/t ^{4})\] s''(2)= -0.25 m/s^2
@archanatikayatray its actually 4/t^3-12/t^4 Thank you!
@arshia93 You're Welcome
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