The given velocity of the particle moving in the xy plane is:
v = (6.0t - 4.0t^2) i^ + 8.0 j^
(a) To find the acceleration when t = 3.0s, we differentiate the velocity with respect to time:
a = dv/dt = (6.0 - 8.0t) i^
Substituting t = 3.0s, we get:
a = (6.0 - 8.0(3.0)) i^ = -18.0 i^
Therefore, the acceleration when t = 3.0s is -18.0 m/s^2 in the x-direction.
(b) To find when the acceleration is zero, we set the acceleration to zero and solve for t:
a = (6.0 - 8.0t) i^ = 0
Solving for t, we get:
t = 0.75 seconds
Therefore, the acceleration is zero when t = 0.75 seconds.
(c) To find when the velocity is zero, we set the velocity to zero and solve for t:
v = (6.0t - 4.0t^2) i^ + 8.0 j^ = 0
Solving for t, we get:
t = 0 seconds and t = 1.5 seconds
Therefore, the velocity is zero at t = 0 seconds and t = 1.5 seconds.
(d) To find when the speed equals 10 m/s, we first need to find the magnitude of the velocity:
|v| = sqrt((6.0t - 4.0t^2)^2 + 8.0^2)
Setting this equal to 10 m/s and solving for t, we get:
t = 0.981 seconds and t = 2.019 seconds
Therefore, the speed is equal to 10 m/s at t = 0.981 seconds and t = 2.019 seconds.