The point moves in a straight line according to the law x (t) = 2t ^ 3 + 3t +1. Find its acceleration at time t = 3c.

In order to find the acceleration a (t), you need to find the second derivative of the law of motion with respect to the variable t.

First, we find the speed V (t) = x ‘(t) = (2t ^ 3 + 3t + 1)’ = 2 * 3 * t ^ 2 + 3 + 0 = 6t ^ 2 + 3.

Now we can find the acceleration a (t) = V ‘(t) = x’ ‘(t) = (6t ^ 2 + 3)’ = 6 * 2 * t + 0 = 12t.

a (3) = 12 * 3 = 36 m / s ^ 2.

Answer: the acceleration of a point at time t = 3c is a (3) = 36 m / s ^ 2.