1. On the slope with a sine value of 1 / 10, a car runs with constant power, and the friction resistance is 2 / 10 of the weight of the car. If the speed rate of the car going up the slope at a constant speed is V, what is the speed rate of the car going down the slope at a constant speed? 2. The vehicle accelerates on the horizontal road with constant power and constant resistance. When the speed is V, the acceleration is a, and when the speed is 2V, the acceleration is a / 4, what is the maximum speed of the vehicle?

1. On the slope with a sine value of 1 / 10, a car runs with constant power, and the friction resistance is 2 / 10 of the weight of the car. If the speed rate of the car going up the slope at a constant speed is V, what is the speed rate of the car going down the slope at a constant speed? 2. The vehicle accelerates on the horizontal road with constant power and constant resistance. When the speed is V, the acceleration is a, and when the speed is 2V, the acceleration is a / 4, what is the maximum speed of the vehicle?

When going uphill at a constant speed, the traction force F = P / V, which is equal to mgsin α + F = mg / 10 + 2mg / 10 = 3mg / 10, so the power P = 3mgv / 10. When going downhill at a constant speed, the traction force plus gravity component is equal to the friction resistance. F + mg / 10 = 2mg / 10, so f = mg / 10. And P = FV, so v = 3m / s, when the speed is V, the traction force F = P / v. P / V-F = ma, when the speed is v

The mass of the car is m, and the constant power of the engine is p. when he drives on a slope with an inclination angle of a, the resistance is k times of the weight of the car, and the maximum speed of the car is calculated

P = f traction * V, when V is the maximum, f traction = (K + Sina) mg. From these two formulas, the fastest speed v = P / ((K + Sina) mg) can be obtained

A car moving along a slightly inclined slope, if the engine rate remains unchanged, it can go up the slope at V1 speed and go down the slope at V2 speed, then it can move at a constant speed on the same roughness of the horizontal road for about 20 minutes——————
A,√（V1V2)
B,(V1+V2)/2
C,2V1V2/(V1+V2)
D,V1V2/(V1-V2)

P = (f friction + F gravity component) V1 = (f friction - f gravity component) V2
2 F friction = P / V1 + P / V2
V = P / F friction = 2v1v2 / (V1 + V2)
C

When a car moves along a micro slope, the inclination angle of the slope is set as θ. If the power of the engine is kept constant, it can go up the slope at a constant speed of V1 and go down the slope at a constant speed of V2, what is its speed on the horizontal road with the same roughness as the slope? (constant power)

If the power is constant and the motion is uniform, the traction is equal to the resistance
When uphill, P = f1v1 = (μ mgcos θ + mgsin θ) v1
When going downhill, P = f2v2 = (μ mgcos θ - mgsin θ) V2
The solution is μ = (V1 + V2) Tan θ / (v2-v1)
On the horizontal road
P=μmgV
V=2V1V2cosθ/(V1+V2)