Physics, 01.02.2021 23:10 sharonkrobinson

The process of braking or accelerating an automobile in heavy traffic is highly complex, requiring the skillful use of both feedback and feedforward mechanisms to drive safely. Consider a controlled variable as the distance between you and the car in front of you, with some specified distance as a set point. Name one feedback and one feedforward control mechanisms would keep you at that distance, and prevent you from colliding with the car in front of you

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A feedback mechanism is the constant measurement of the distance between the two vehicles and with this the calculation of the speed between them

An anticipated control mechanism is using the vehicle's acceleration and its deceleration to calculate the future speed and their distances,

Explanation:

For this exercise, the feedback and control mechanisms must be directly related to the kinematic relationships,

It is assumed that the vehicle speed (taken from the speedometer) and the braking capacity (given by the brake manufacturer) are known in the form of negative acceleration,

A feedback mechanism is the constant measurement of the distance between the two vehicles and with this the calculation of the speed between them, for which we know the acceleration that exists. This would be a correct mechanism, in general we can adjust to an error between the sedated distance and the real one, so when they have very different give a maximum acceleration and in decreasing it as the differences between the distances decrease.

An anticipated control mechanism is using the vehicle's acceleration and its deceleration to calculate the future speed and their distances, so we would know the amount of acceleration necessary to reach the optimal distance between the two vehicles.

Answer from: Quest

double the distance = divide strength by 4

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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