Using the coordinate equation x = 6 + 2t + 6t ^ 2, determine: initial speed, projection of acceleration, initial coordinate

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

x0 -?

V0 -?

a – ?

1 way.

x0 = x (0) = 6 + 2 * 0 + 6 * 0 ^ 2 = 6 m.

Velocity is the first derivative of the coordinate: V (t) = x (t) “.

V (t) = (6 + 2 * t + 6 * t ^ 2) “= 2 + 12 * t.

V0 = V (0) = 2 + 12 * 0 = 2 m / s.

The acceleration of the body is the second derivative of the coordinate: Vа (t) = х (t) “”

a (t) = (2 + 12 * t) “= 12 m / s2.

Method 2.

The dependence of the coordinate on time for uniformly accelerated motion has the form: x (t) = x0 + V0 * t + a * t ^ 2.

x0 = 6 m, V0 = 2 m / s, a = 12 m / s2.

Answer: x0 = 6 m, V0 = 2 m / s, a = 12 m / s2.