- #1
laker_gurl3
- 94
- 0
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