The point moves in a straight line according to the law x = 6t – t². At what point will its speed become zero?

Given:

x = 6t – t ^ 2

v (t) = 0

t-?

We know the formula: x = x (start) + v (start) * t + (a * t ^ 2) / 2, where x (start) is the initial coordinate of the point, v (start) is the initial speed of the point, and is the acceleration points.
According to this formula, our point has the following parameters:
x (start) = 0

v (start) = 6 m / s

a = -2 m / s ^ 2.

3. We know the formula t = (v – v (start)) / a.

4. Substitute: t = (0 -6) / (-2) = 3c.

Answer: t = 3s.