A particle of mass 2 kg starts from the origin at t=0 . Its velocity vector as a function of time is given by v (t) = (3t^2 i + 4t j ) m/s . The net force vector acting on the particle at the exact instant its x -coordinate reaches 8 m is:

Step-by-step solution

Given the velocity vector v (t) = 3t^2 i + 4t j , the x -component of velocity is v_x = 3t^2 . The x -coordinate as a function of time is found by integrating v_x with respect to time: x(t) = 3t^2 dt = t^3 (since it starts from the origin, the constant of integration is zero). We are given that the target x -coordinate is 8 m . Setting x(t) = 8 gives: t^3 = 8 t = 2 s . Next, we find the acceleration vector by differentiating the velocity vector with respect to time: a (t) = d v dt = 6t i + 4 j At t = 2 s , the acce