Question

physics {deceleration}

Answer 1

Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)