A particle is moving in the with acceleration (2t - 3) ms^-2 and initial velocity 2ms^-1. Find the distance travelled when the velocity has reached 12ms^-1.

(1.) Integrate the expression for acceleration to find an expression for velocity: Velocity v = t^2 - 3t + c        When t = 0, velocity = 2. Substituting in to find constant c, 2 = 0 + 0  + c therefore c = 2.    v = t^2 - 3t + 2       (2.) Find the time taken for velocity to reach 12ms^-1.     t^2 - 3t + 2 = 12   so   t^2 -3t - 10 = 0. Factorising gives (t-5)(t+2)=0, so t = 5 or t = -2. Since time must be positive, t = 5.  (3.) Integrate the expression for velocity to find an expression for distance travelled.  Displacement x = (1/3)t^3 - (3/2)t^2 + 2t +d         When t = 0, displacement = 0 therefore d = 0.     x = (1/3)t^3 - (3/2)t^2 + 2t (4.) Find displacement when t = 5.    When t = 5, x = 85/6 metres.  Hence distance travelled = 85/6 m, approximately 14.2m.