Question

The acceleration of a particle moving only on a horizontal xy plane is given by , where is in meters per second-squared and t is in seconds. at t = 0, the position vector locates the paticle, which then has the velocity vector . at t = 2.70 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?

Answer 1

Integrate a to get v and use initial data for v to evaluate the constants of integration. v = [(3/2)t^2 + Ci]i + [2t^2 + Cj]j When t=0, v = 5i + 2j, hence Ci=5, Cj=2 So v = [(3/2)t^2 + 5]i + [2t^2 + 2]j <==ANS (b) Integrate v to get r and use initial data for r to evaluate the constants of integration: The result is: r = [0.5t^3 + 5t + 20]i + [(2/3)t^3 + 2t + 40]j <==ANS Set t=4 in the above expression to evaluate (c) Convert the result of (c) to polar form to evaluate (d)