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y''= -3/(t^3)
s=0 when t = 1
s'= -3/2 when t= (3)^1/2
my answer is....
-3/(2t^-2) - 2t + 7/2
please tell me if that is correct
s=0 when t = 1
s'= -3/2 when t= (3)^1/2
my answer is....
-3/(2t^-2) - 2t + 7/2
please tell me if that is correct