Final answer:
At t = 3 s, the particle has a position of r(3) = 60i + 108j, a velocity of v(3) = 20i + 72j, and a constant acceleration of a = 24j m/s².
Step-by-step explanation:
The question relates to kinematics in Physics, concerning a particle's position function given as r(t) = 20ti + 12t². To find the position, velocity, and acceleration at a specific time (t=3 s), we first need to evaluate r(3) by plugging in t=3 into the position function, v(t) as the derivative of r(t) to get the velocity, and a(t) as the derivative of v(t) for acceleration.
To find the position at t = 3 s, substitute t=3 into r(t). The velocity v(t) is the first derivative of r(t), v(t) = 20i + 24t, and acceleration a(t) is the second derivative of r(t), a(t) = 24.
Therefore, at t = 3 s, the position is r(3) = 20(3)i + 12(3)² = 60i + 108 = 60i + 108j, the velocity is v(3) = 20i + 24(3) = 20i + 72j, and the acceleration remains constant at a = 24j m/s².