calculus
posted by Preet .
The displacement (cm) of a point is given by s= 〖4t〗^33√(1+2t) . Find the instantaneous velocity and acceleration at t = 1.5 sec.
Round to 3 significant figures

s = (4t)^3  3√(1+2t)
ds/dt = 3(4t)^2 * 4  3/√(1+2t)
= 12(4t)^2  3/√(1+2t)
d2s/dt2 = 12*2(4t)*4 + 3/√(1+2t)^3
= 96(4t) + 3/√(1+2t)^3
Now just plug in t = 3/2 and evaluate
s(3/2) = 6^3  3√4 = 2166 = 210
s'(3/2) = 12(6^2)  3/√4 = 4323/2 = 430.5
s''(3/2) = 96(6) + 3/√4^3 = 576+3/8 = 576.375