Question

Suppose that the position of a particle is given by s=f(t)=2 t³ +7 t+9

(a) Find the velocity at time t). v(t)={m}{s}
(b) Find the velocity at time t=3 seconds.

Answer 1

In Physics, the velocity at time t for the position function s=f(t)=2 t³ +7 t+9 is the derivative of the function, which is 6t² + 7 m/s. The specific velocity at t=3 seconds is found to be 61 m/s when you substitute t=3 into the velocity function.

Step-by-step explanation:

The student is dealing with a problem that involves calculus applied to kinematics, a branch of Physics. Specifically, this question requires finding the velocity, which is the derivative of the position function s=f(t)=2 t³ +7 t+9. To answer part (a) of the question, we find the first derivative of the position function with respect to time which gives us the velocity function v(t). For part (b), we evaluate the velocity function at t=3 seconds.

Step-by-step explanation:

Calculate the velocity function, v(t), by differentiating the position function:
v(t) = s'(t) = d/dt(2t³ + 7t + 9) = 6t² + 7.

Find the velocity at time t=3 seconds:
v(3) = 6(3)² + 7 = 54 + 7 = 61 m/s.

The velocity at time t is v(t) = 6t² + 7 m/s, and at t=3 seconds, the velocity is 61 m/s.