Velocity of particle at time t=0 is 2m/s. A constant acceleration of 2m/s^2 acts on the particle for 1 sec at angle of 60° with it`s initial velocity. find the magnitude of velocity at the end of 1 second.

Direction along the initial velocity: Vo = 2m/s, a = 2m/s^2 * cos(60) = 1m/s^2, t = 2 second Use Vf = Vo + a * t = 2m/s + 1m/s^2 * 2s = 4m/s. Use S = Vo*t + (1/2) * a * t^2 = 2m/s * 2s + (1/2) * 1m/s^2 * (2s)^2 = 4m + 2m = 6m Direction perpendicular to the initial velocity: Vo =0m/s, a = 2m/s^2 * sin(60) = 1.732m/s^2, t = 2 second Again, Use Vf = Vo + a * t = 0 + 1.732m/s^2 * 2s = 3.46m/s. Use S = Vo*t + (1/2) * a * t^2 = 0 + (1/2) *1.732m/s^2 * (2s)^2 = 3.46m Now, combine velocities in both direction (Vectorially) , and combine diplacement in both direction (Vectorially). This left for you to be done.