The mass is 3100 kg, the speed is 15 m / s, the friction force is 45000N when braking, the speed

The mass is 3100 kg, the speed is 15 m / s, the friction force is 45000N when braking, the speed is 0, find the braking distance.

Initial data: m (conditional body weight) = 3100 kg; V0 (speed before braking) = 15 m / s; Ftr. (acting frictional force) = 45000 N; V (speed of the conditional body after deceleration) = 0 m / s.

1) Acceleration of the conditional body during deceleration: F = -Ftr. = m * a, whence a = -Ftr. / m = -45000 / 3100 = -14.52 m / s2.

2) Braking distance: S = (V ^ 2 – V0 ^ 2) / 2a = -V0 ^ 2 / 2a = -15 ^ 2 / (2 * (-14.52)) = 7.75 m.

Answer: The stopping distance of the conventional body is 7.75 m.