V,? = V2 + 2a Ax
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Answer:
The shortest possible stopping distance of the car is 175.319 meters.
Explanation:
In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:
(Eq. 1)
(Eq. 2)
After quick handling, we get that deceleration experimented by the car is equal to:
(Eq. 3)
Where:
- Deceleration of the car, measured in meters per square second.
- Kinetic coefficient of friction, dimensionless.
- Gravitational acceleration, measured in meters per square second.
If we know that and
, then deceleration of the car is:
The stopping distance of the car (), measured in meters, is determined from the following kinematic expression:
(Eq. 4)
Where:
- Initial speed of the car, measured in meters per second.
- Final speed of the car, measured in meters per second.
If we know that ,
and
, stopping distance of the car is:
The shortest possible stopping distance of the car is 175.319 meters.
Answer:
Explanation:
The relationship between the refractive index and the critical angle is given as follows:
where,
η = refractive index = 1.67
θc = critical angle =?
Therefore,