Question

answer the ninth question

Answer 1

s = t^3/3

a = d2s/dt2 = d/dt (t^2) = 2t
m = 3kg
F = ma =3×2t = 6t

ds = d(t^3/3) = t^2 dt

Work done = integral {F.ds}

W = integral(0 to 2) {6t . t^2. dt}

W = integral(0 to 2) {6t^3 dt}

W = [1.5 t^4] (from 0 to 2) = 1.5×(2^4 - 0) = 24J


Answer is D.
Hope it helps.
Mark as brainliest.

Answer 2

Answer:

given mass (m)=3kg

Given S=t³/3(in m)

          a=$\frac{d^2 s}{dt^2}$

            =d(3t/3)dt=d(t)/dt=2t

  F=ma=3(2t)=6t

Work done (W)=$\int\limits^{(t =2)}_{(t=0)} {F} \, ds$

                          =int at t=2{6t.t²}ds

                             =int at t=2 (6t³)ds

                                =6(int t³)

                               =24J

hope this helps you

mark me as the brainliest