Given the equation of motion of the body x = 6 + 4t + t^2 find the initial coordinate of the body

Given the equation of motion of the body x = 6 + 4t + t^2 find the initial coordinate of the body, initial speed, acceleration, equation of speed, equation of displacement, nature of motion.

x (t) = 6 + 4 * t + t ^ 2.

x0 -?

V0 -?

V (t) -?

S (t) -?

With uniformly accelerated motion, the coordinate of the body changes according to the law: x (t) = x0 + V0 * t + a * t ^ 2/2, where x0 is the initial velocity of the body, V0 is the initial velocity of the body, and is the acceleration during uniformly accelerated motion.

For the dependence x (t) = 6 + 4 * t + t ^ 2, the body moves rectilinearly from the initial coordinate x0 = 6 m, with the initial speed of movement V0 = 4 m / s, with the acceleration of motion a = 2 m / s2.

The speed of the body changes according to the law: V (t) = V0 + a * t.

V (t) = 4 + 2 * t.

S (t) = x (t) – x0 = x0 + V0 * t + a * t ^ 2/2 – x0 = V0 * t + a * t ^ 2/2.

Answer: x0 = 6 m, V0 = 4 m / s, a = 2 m / s2, V (t) = 4 + 2 * t, S (t) = V0 * t + a * t2 / 2.