# At the start of observation, the distance between the two bodies is 6.9 m. The first body moves from a state

At the start of observation, the distance between the two bodies is 6.9 m. The first body moves from a state of rest with an acceleration of 0.2 m / s2. The second moves after him, having an initial speed of 2 m / s and an acceleration of 0.4 m / s2. Write the equations x = x (t) in the frame of reference, in which at t = 0 the coordinates of the bodies take on values correspondingly equal to x1 = 6.9 m, x2 = 0. Find the time and place of meeting of the bodies.

S = 6.9 m.

V01 = 0 m / s.

a1 = 0.2 m / s2.

X01 = 6.9 m.

V02 = 2 m / s.

X02 = 0 m.

a2 = 0.4 m / s2.

X1 (t) -?

X2 (t) -?

Hvst -?

tvst -?

In rectilinear motion with constant acceleration, the dependence of the body coordinate on time X (t) has the form: X (t) = X0 + V0 * t + a * t2 / 2, where X0 is the initial coordinate of the body, V0 is the initial velocity of the body, t is the time of movement, and is the acceleration of the body.

X1 (t) = X01 + V01 * t + a1 * t2 / 2.

X1 (t) = 6.9 + 0.2 * t2 / 2 = 6.9 + 0.1 * t2.

X2 (t) = X02 + V02 * t + a2 * t2 / 2.

X2 (t) = 2 * t + 0.4 * t2 / 2 = 2 * t + 0.2 * t2.

The bodies will meet when they have the same coordinate: X1 (t) = X2 (t).

6.9 + 0.1 * t2 = 2 * t + 0.2 * t2.

0.1 * t2 + 2 * t – 6.9 = 0.

t2 + 20 * t – 69 = 0.

D = 202 – 4 * 1 * (- 69) = 400 + 276 = 676.

t1,2 = (- 20 ± √676) / 2.

t1 = 3, t2 = -23 – the root is meaningless.

tvst = 3 s.

Hvst = 6.9 +0.1 * (3) 2 = 7.8 m.

Answer: twst = 3 s, Hvst = 7.8 m.

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