The kinematic law of motion of a material point along the Ox axis has the form: x = A + Bt + Ct2

The kinematic law of motion of a material point along the Ox axis has the form: x = A + Bt + Ct2, where A = 2.00 m, B = –1.50 m / s, C = 2.00 m / s2. Write down the equation of the dependence of the projection of the speed of movement of a material point on time. Determine the coordinate of the point and the projection of the speed of its movement through the time interval Δt1 = 6.00 s from the beginning of the movement.

x (t) = A + B * t + C * t ^ 2.

A = 2.00 m.

B = -1.50 m / s.

C = 2.00 m / s2.

t1 = 6.00 s.

Vx (t) -?

x1 -?

V1x -?

With uniformly accelerated motion, the coordinate of the body has a time dependence of the form: x (t) = x0 + V0x * t + ax * t ^ 2/2, where x0 is the initial coordinate of the body, V0x is the projection of the initial velocity, and ax is the projection of acceleration.

x (t) = A + B * t + C * t ^ 2, then the initial coordinate of the body is x0 = A, the projection of the initial velocity is V0x = B, the projection of the acceleration is ax = 2 * C.

x1 = 2.00 m – 1.50 m / s * 6.00 s + 2.00 m / s2 * (6.00) ^ 2 = 65 m.

With uniformly accelerated motion, the projection of the velocity Vx (t) changes according to the law: Vx (t) = V0x + ax * t.

Vx (t) = B + 2 * C * t.

Vx (t) = – 1.50 m / s + 2 * 2.00 m / s2 * t.

V1x = – 1.50 m / s + 2 * 2.00 m / s2 * 6.00 s = 22.50 m / s.

Answer: x1 = 65 m, V1x = 22.50 m / s, Vx (t) = – 1.50 m / s + 2 * 2.00 m / s2 * t.