After the traction of the locomotive stopped, the train stopped on a horizontal section after 60 s.

After the traction of the locomotive stopped, the train stopped on a horizontal section after 60 s. Determine the distance traveled by the train during this time if it is known that the coefficient of resistance to movement is equal to 0.4.

t = 60 s.

V = 0 m / s.

g = 9.8 m / s2.

μ = 0.4.

S -?

Let’s write Newton’s 2 law for a locomotive after turning off the engines: Ftr = m * a.

The friction force Ftr is determined by the formula: Ftr = μ * m * g, where μ is the coefficient of friction, m * g is the force of gravity.

μ * m * g = m * a.

When braking, the locomotive will move with acceleration a = μ * g.

a = 0.4 * 9.8 m / s2 = 3.92 m / s2.

We express the acceleration by the kinematic formula: a = (V0 – V) / t.

Since the locomotive has stopped, the final speed will be 0, V = 0 m / s.

a = V0 / t.

V0 = a * t.

V0 = 3.92 m / s2 * 60 s = 235.2 m / s.

S = V02 / 2 * a.

S = (235.2 m / s) 2/2 * 3.92 m / s2 = 7056 m.

Answer: the braking distance of the locomotive was S = 7056 m.