A particle starts from the origin at t=0 with an initial velocity of (4 i + 2 j ) m s ⁻¹ and moves in the x-y plane with a constant acceleration of (- i - 3 j ) m s ⁻² . The square of its distance from the origin (in m ^2 ) at the instant its velocity vector becomes perpendicular to its initial velocity vector is _______.

Step-by-step solution

The initial velocity vector is u = 4 i + 2 j . The acceleration vector is a = - i - 3 j . The velocity vector as a function of time t is: v (t) = u + a t = (4 - t) i + (2 - 3t) j For the velocity vector to be perpendicular to the initial velocity, their dot product must be zero: v (t) u = 0 [(4 - t) i + (2 - 3t) j ] (4 i + 2 j ) = 0 4(4 - t) + 2(2 - 3t) = 0 16 - 4t + 4 - 6t = 0 20 - 10t = 0 t = 2 s Now, we calculate the coordinates of the particle at t = 2 s using s = u t + 1 2 a t^2 : x = u_x t + 1 2 a_x t^2 = 4(2