Final answer:
The velocity of a particle with the position function s(t) = 6t³ + 7t + 9 is v(t) = 18t² + 7 m/s. At t = 3 seconds, the velocity is 169 m/s. The acceleration is a(t) = 36t m/s², and at t = 3 seconds, the acceleration is 108 m/s².
Step-by-step explanation:
To find the velocity of a particle when its position is given as a function of time, we calculate the first derivative of the position function with respect to time. For the position function s(t) = 6t³ + 7t + 9, the velocity function v(t) is found by differentiating the position function:
v(t) = ds/dt = d/dt (6t³ + 7t + 9) = 18t² + 7.
To find the velocity at time t = 3 seconds, we substitute t with 3:
v(3) = 18(3)² + 7 = 18(9) + 7 = 162 + 7 = 169 m/s.
To find the acceleration, we differentiate the velocity function:
a(t) = dv/dt = d/dt (18t² + 7) = 36t.
For the acceleration at time t = 3 seconds:
a(3) = 36(3) = 108 m/s².