The point moves according to the law S = 2 t ^ 3-2t + 5. Find the speed and acceleration of the movement at t = 3c

Velocity is the first derivative of distance, and acceleration is the second derivative of it. Therefore, if the equation of motion has the form S (t) = 2t³ – 2t + 5, then the equation of velocity versus time will have the form:

V (t) = S ‘= (2t³ – 2t + 5)’ = 2 * 3 * t² – 2 + 0 = 6t² – 2.

And the equation of acceleration versus time:

a (t) = S ” = V ‘= (6t² – 2)’ = 6 * 2 * t – 0 = 12t.

Then at time t = 3 s:

V (3) = 6 * 3² – 2 = 6 * 9 – 2 = 36 – 2 = 34 m / s.

a (3) = 12 * 3 = 36 m / s².

Answer: at the moment of time 3 s the speed is 34 m / s, and the acceleration is 36 m / s².