A particle, of mass 400 grams, is initially at rest at the point O. The particle starts to move in a straight line so that its velocity, vms^-1 at time I seconds is given by v=6t^2-12t^3 for tgt 0 (a) Find an expression in terms of I for the force acting on the particle. (b) Find the time when the particle next passes through O. (3) (5)

## Step1:Firstly, we know from Newton’s second law of motion, the Force exerted on an object is the product of its mass and acceleration, usually expressed as . Here in the question, acceleration 'a' can be represented as the derivation of velocity 'v' with respect to time 't'. ### ## Step2:We derive the velocity function given in the question, with respect to time 't'. The derivation of is , and the derivation of is .Using chain rules, derive the above expression. ### ## Step3:Substitute the derived and the given m=400grams=0.4 kg into the Newton’s second law equation of `Step 1`.### \(F =m \cdot \frac{ dv }{ dt }= 0.4( 12t - 36t^{2} ) = 4.8t - 14.4t^{2} \) For clear understanding & for simplifying fractions we may also represent 18 as 4.8 and later recognizing 2.88 in mathematics. ## Step4: Understand that the particle passes through O again when the displacement which is determined by the integral of velocity ( ) equals to zero, considering that the particle initially positions at O. This will happen when the velocity cancels out the movement, that is, resulting velocity \(v = \int vd(t) = 0 \).## Step5:Solve the integral \(\int vd(t) = \int (6t^{2} - 12t^{3})\ dt \)###