A particle starts from the origin at t=0 with an initial velocity of 3 j m s ⁻¹ . It moves in the x-y plane under the action of a constant acceleration of (4 i + c j ) m s ⁻² , where c is a positive constant. At the instant the x -coordinate of the particle is 18 m , its speed is 15 m s ⁻¹ . The value of c is _______.

Step-by-step solution

Given, u_x = 0 , u_y = 3 m s ⁻¹ a_x = 4 m s ⁻² , a_y = c At a certain instant, x = 18 m . Using the second equation of motion for the x -direction: x = u_x t + 1 2 a_x t^2 18 = 0 + 1 2 (4)t^2 2t^2 = 18 t^2 = 9 t = 3 s The velocity of the particle along the x -axis at t = 3 s is: v_x = u_x + a_x t = 0 + 4(3) = 12 m s ⁻¹ The total speed is given as v = 15 m s ⁻¹ . Since v^2 = v_x^2 + v_y^2 , we have: 15^2 = 12^2 + v_y^2 225 = 144 + v_y^2 v_y^2 = 81 v_y = 9 m s ⁻¹ Using the first equation of motion for the y -directio