The velocity of a particle is given as v = -x i + 2y j - z k m/s. The magnitude of acceleration at point (1, 2, 4) is _______ m/s ^2 .

Step-by-step solution

The velocity vector is given by v = v_x i + v_y j + v_z k = -x i + 2y j - z k . The components of acceleration are: a_x = v_x v_x x + v_y v_x y + v_z v_x z = (-x)(-1) + 0 + 0 = x a_y = v_x v_y x + v_y v_y y + v_z v_y z = 0 + (2y)(2) + 0 = 4y a_z = v_x v_z x + v_y v_z y + v_z v_z z = 0 + 0 + (-z)(-1) = z The acceleration vector is a = x i + 4y j + z k . At the point (1, 2, 4) , substituting x = 1 , y = 2 , and z = 4 : a = 1 i + 4(2) j + 4 k = i + 8 j + 4 k The magnitude of acceleration is: | a | = 1^2 + 8^2 + 4^2 =