a = 24-15√t
v = 24t-10t^(3/2)
s = 12t^2 - 4t^(5/2)
s=0 when 12-4√t = 0, or t=9
a(9) = 24-15√9 = -21
a particle starts from rest and moves along a straight line with an acceleration equal to 24-15√t where distance and time are measured in terms of feet and seconds respectively. with what acceleration will the particle return to the starting point?
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