Find the velocity vector v(t) and acceleration vector a(t).

Step 1: Find the velocity vector v(t) and acceleration vector a(t). The position vector of particle P is given by r = [(18t - 4t^3)i + t^2j] m. The mass of the particle is m = 6 kg. To find the velocity vector v(t), differentiate the position vector r(t) with respect to time t: v(t) = (dr)/(dt) = (d)/(dt)((18t - 4t^3)i + t^2j) v(t) = (18 - 12t^2)i + (2t)j m/s To find the acceleration vector a(t), differentiate the velocity vector v(t) with respect to time t: a(t) = (dv)/(dt) = (d)/(dt)((18 - 12t^2)i + (2t)j) a(t) = (-24t)i + (2)j m/s^2 a) Find the kinetic energy of P when t = 1. First, find the velocity of P at t=1 s: v(1) = (18 - 12(1)^2)i + (2(1))j v(1) = (18 - 12)i + 2j v(1) = 6i + 2j m/s Next, calculate the magnitude of the velocity |v(1)|: |v(1)| = sqrt(6^2 + 2^2) = sqrt(36 + 4) = sqrt(40) m/s The kinetic energy KE is given by the formula KE = (1)/(2)mv^2. KE = (1)/(2)(6 kg)(sqrt(40) m/s)^2 KE = (1)/(2)(6)(40) KE = 3 × 40 KE = 120 J The kinetic energy of P when t=1 is 120 J. b) Calculate the magnitude of the force acting on P when t = 2. First, find the acceleration of P at t=2 s: a(2) = (-24(2))i + (2)j a(2) = -48i + 2j m/s^2 Next, calculate the force vector F(2) using Newton's second law, F = ma: F(2) = (6 kg)(-48i + 2j m/s^2) F(2) = -288i + 12j N Finally, calculate the magnitude of the force |F(2)|: |F(2)| = sqrt((-288)^2 + 12^2) |F(2)| = sqrt(82944 + 144) |F(2)| = sqrt(83088) N |F(2)| ≈ 288.250 N The magnitude of the force acting on P when t=2 is sqrt(83088) N ≈ 288.25 N. c) Determine, to 2 decimal places, the value of t for which the velocity and the acceleration vectors of P are at right angles. If two vectors are at right angles, their dot product is zero. So, $v(t) · a(t) =