This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here is the solution to question 1(i): 1(i) A force F = (18i + 18j - 36k) N acts at time t seconds on a particle of mass 3kg. Initially, the particle is at the origin moving with velocity (2i - j - 3k) ms^-1. a) Find the acceleration of the particle at time t. Step 1: Use Newton's second law, F = ma. Given force F = (18i + 18j - 36k) N and mass m = 3 kg. a = Fm Step 2: Substitute the given values. a = (1)/(3)(18i + 18j - 36k) Step 3: Simplify to find the acceleration. a = (6i + 6j - 12k)\, ms^-2 The acceleration of the particle is (6i + 6j - 12k)\, ms^-2. b) Find the velocity of the particle at time t. Step 1: Use the kinematic equation for velocity, v = u + at. Given initial velocity u = (2i - j - 3k) ms^-1 and acceleration a = (6i + 6j - 12k) ms^-2 (from part a). Step 2: Substitute the values into the equation. v = (2i - j - 3k) + (6i + 6j - 12k)t Step 3: Group the components. v = (2+6t)i + (-1+6t)j + (-3-12t)k\, ms^-1 The velocity of the particle at time t is (2+6t)i + (-1+6t)j + (-3-12t)k\, ms^-1. c) When t=2, find the kinetic energy of the particle. Step 1: Find the velocity vector at t=2 s using the expression from part b. v(2) = (2+6(2))i + (-1+6(2))j + (-3-12(2))k v(2) = (2+12)i + (-1+12)j + (-3-24)k v(2) = 14i + 11j - 27k\, ms^-1 Step 2: Calculate the magnitude of the velocity vector, |v| = sqrt(v_x^2 + v_y^2 + v_z^2). |v(2)| = sqrt(14^2 + 11^2 + (-27)^2) |v(2)| = sqrt(196 + 121 + 729) |v(2)| = sqrt(1046)\, m/s Step 3: Calculate the kinetic energy using the formula KE = (1)/(2)mv^2. Given mass m = 3 kg. KE = (1)/(2)(3)(sqrt(1046))^2 KE = (3)/(2)(1046) KE = 3 × 523 KE = 1569\, J The kinetic energy of the particle at t=2 s is 1569\, J. d) A particle starts from rest and moves with an acceleration of (2v^2+1) ms^-2 where v is the speed of the particle. Find the distance taken by the particle to attain a speed of 3m/s. Step 1: Relate acceleration to velocity and distance using a = v(dv)/(dx). Given acceleration a = 2v^2+1. v(dv)/(dx) = 2v^2+1 Step 2: Separate variables and integrate. (v)/(2v^2+1)\,dv = dx Integrate from initial conditions (v=0 at x=0) to final conditions (v=3 at x). _0^3 (v)/(2v^2+1)\,dv = _0^x dx Step 3: Evaluate the integral. For the left side, let u = 2v^2+1, so du = 4v\,dv v\,dv = (1)/(4)\,du. When v=0, u = 2(0)^2+1 = 1. When v=3, u = 2(3)^2+1 = 18+1 = 19. _1^19 (1)/(4u)\,du = [x]_0^x (1)/(4)[|u|]_1^19 = x