When the motorcycle was driving on a horizontal highway at a speed of 54 km / h, the driver turned

When the motorcycle was driving on a horizontal highway at a speed of 54 km / h, the driver turned off the engine. after that, the motorcycle moved to a stop with equal slowness to find the acceleration path and time from off. engine if the coefficient of friction of wheels on the road is 0.03

1) Let us find the resultant of all forces acting on the motorcycle according to Newton’s second law: ma = Fthrust – F friction. Since the driver has turned off the engine, Fthroat = 0 N. And the friction force is found through the reaction force of the support N = mg; F friction = µ • N = µ • mg. The friction coefficient is µ = 0.03, g = 9.8 N / kg, and the body weight is m. In order to determine what the acceleration is equal to, we express it from Newton’s law: a = (Fthrust – µ • mg): m; a = (0 – µ • mg): m; a = – µ • g. Substitute the known values: a = – 0.03 • 9.8 m / (s • s). = – 0.294 m / (s • s).
2) After turning off the engine, the motorcycle moved with equal slowness until it stopped, changing its speed from Vo = 54 km / h = 15 m / s to V = 0 m / s. In order to determine the braking time, we use the definition of acceleration a = Δv: t, we express t = Δv: a, where the change in speed is Δv = V – Vо m / s; Δv = 0 m / s – 15 m / s = – 15 m / s, then the time during which this change in speed occurs t = – 15 m / s: (- 0.294 m / (s • s)) = 51 s.
3) To find the braking distance, we use the formula for the path for uniformly accelerated movement: S = (V + Vо) • t / 2; then S = (0 m / s + 15 m / s) • 51s / 2; S = 382 m.
Answer: when braking a = – 0.294 m / (s • s); t = 51s; S = 382 m.