It is stipulated that the speed of a vehicle should not exceed 70 km / h on a certain highway. When a traffic accident occurs, the traffic police usually estimate the speed of the vehicle according to the sliding distance of the wheel after braking. The empirical formula used is v = 16 √ DF, where V is the vehicle speed (unit: km / h), D is the sliding distance of the wheel after braking (unit: m), and F is the friction factor, D = 20m, f = 1.2m, did the speed of the car exceed the specified speed?

It is stipulated that the speed of a vehicle should not exceed 70 km / h on a certain highway. When a traffic accident occurs, the traffic police usually estimate the speed of the vehicle according to the sliding distance of the wheel after braking. The empirical formula used is v = 16 √ DF, where V is the vehicle speed (unit: km / h), D is the sliding distance of the wheel after braking (unit: m), and F is the friction factor, D = 20m, f = 1.2m, did the speed of the car exceed the specified speed?

V = 16 √ (20 × 1.2) = 16 √ 24, about 78 km / h
So it's more than that

If the throttle is not turned off and the power remains unchanged, the speed in TS will increase by two times, and the initial speed of the vehicle will be zero___ ．

When you turn off the throttle when you go downhill, the speed will not change, indicating that the sliding component of gravity is balanced with the resistance; when you turn on the engine, only the traction can do work when you go downhill. According to the kinetic energy theorem, there is: Pt = 12m (2V) 2-12mv2 solution: v = 2pt3m; so the answer is: 2pt3m

The rated power of a car is p and the mass is m. if the throttle is closed when going downhill, the speed will not change; if the throttle is not closed and the rated power is maintained, the speed will change at t
Then the initial speed of the car is 2 times of the original speed in t time
It's better to have stress analysis,

If the throttle is closed when going downhill, the speed will not change, which means that the downward component of gravity along the inclined plane is equal to the friction force. If the throttle is not closed and the rated power is maintained, the speed will increase by 2 times in t time, which means that the acceleration is made in this process (not uniform acceleration, because the resultant force along the inclined plane is equal to the traction force of the locomotive, and f traction = P / VV is increasing)
Then we get it according to the kinetic energy theorem: the work of combined external force is equal to the change of kinetic energy
PT = 1 / 2mV & # 178; Mo - 1 / 2mV & # 178; initial V Mo = 2V, initial solution is v initial = √ 2pt / 3M

A car with mass m runs at a constant speed on a slope with an inclination angle of θ. The speed is V1 when going up the slope and V2 when going down the slope. If the resistance of the car is constant and the power of the engine is the same when going up and down the slope, the power of the car is equal to______ ．

When uphill, the speed V1, because the uphill process is uniform, the engine traction F1 meets: F1 = mgsin θ + F & nbsp; P = f1v1 = (mgsin θ + F) V1, ① downhill speed V2, because the downhill process is uniform, the engine traction F2 meets: F2 + mgsin θ = F & nbsp; F2 = f-mgsin θ P = f2v2 = (f-mgsin θ) V2, ② from the solution of ① and ②: P = 2mgv1v2sin θ V2 − V1, so the answer is: 2mgv1v2sin θ V2 − v1