The electric tractor moves at a speed of 6.28 km / h. How far the tractor will travel to a complete

The electric tractor moves at a speed of 6.28 km / h. How far the tractor will travel to a complete stop after turning off the engine, if the drag force is 0.3 gravity.

V0 = 6.28 km / h = 1.74 m / s.

V = 0 m / s.

g = 9.8 m / s2.

Fcopr = 0.3 * m * g.

S -?

Let’s write Newton’s 2 law for a tractor after turning off the engine in vector form: m * a = Fcopr + m * g + N.

ОХ: m * a = Fc.

OU: 0 = m * g – N.

m * a = 0.3 * m * g.

a = 0.3 * g.

For uniformly accelerated motion, the path S traversed by the body is found by the formula: S = (V0 ^ 2 – V ^ 2) / 2 * a = (V0 ^ 2 – V ^ 2) / 0.6 * g.

S = ((1.74 m / s) ^ 2 – (0 m / s) ^ 2) / 0.6 * 9.8 m / s2 = 0.51 m.

Answer: the electric tractor will pass the path S = 0.51 m until it stops completely.