A particle moving along the x-axis has its velocity described by the function v =2t squared m/s, where t is in s. Its initial position is x = 2.8m at t = 0 s. At 2.7s , what is the particle's position?

Answer 1

v=dx/dt=2t^2 dx=2t^2dt integrating both side gives: x=2/3t^3+c substituting t=0 and x=2.8 gives c=2.8 therefore x=2/3t^3+2.8 so when t=2.7 x=2/3(2.7)^3+2.8=15.9m