At a particle starts from rest at x=0, y=0 and moves in the xy plane with an acceleration a=(4.0i+3.0j)m/s^2 Determine (a) the x and y components of velocity, (b) the speed of the particle, and (c) the position of the particle, all as a function of time. (d) Evaluate all the above at t=2.0s

(a) The x and y components of velocity can be determined by integrating the acceleration components with respect to time. Since the particle starts from rest, the initial velocity is zero. Therefore, the x and y components of velocity are:<br /><br />$v_x(t) = \int a_x(t) dt = \int 4.0 dt = 4.0t$<br /><br />$v_y(t) = \int a_y(t) dt = \int 3.0 dt = 3.0t$<br /><br />(b) The speed of the particle is given by the magnitude of the velocity vector. It can be calculated as:<br /><br />$r(t) = \sqrt{(v_x(t))^2 + (v_y(t))^2}$<br /><br />$r(t) = \sqrt{(4.0t)^2 + (3.0t)^2}$<br /><br />$r(t) = \sqrt{16.0t^2 + 9.0t^2}$<br /><br />$r(t) = \sqrt{25.0t^2}$<br /><br />$r(t) = 5.0t$<br /><br />(c) The position of the particle can be determined by integrating the velocity components with respect to time. Since the particle starts at the origin, the initial position is zero. Therefore, the position of the particle is:<br /><br />$x(t) = \int v_x(t) dt = \int 4.0t dt = 2.0t^2$<br /><br />$y(t) = \int v_y(t) dt = \int 3.0t dt = 1.5t^2$<br /><br />(d) To evaluate the above expressions at $t=2.0s$, we substitute $t=2.0s$ into the equations:<br /><br />$v_x(2.0s) = 4.0 \times 2.0 = 8.0m/s$<br /><br />$v_y(2.0s) = 3.0 \times 2.0 = 6.0m/s$<br /><br />$r(2.0s) = 5.0 \times 2.0 = 10.0m/s$<br /><br />$x(2.0s) = 2.0 \times (2.0)^2 = 8.0m$<br /><br />$y(2.0s) = 1.5 \times (2.0)^2 = 6.0m$