Answer:
AB = DE <CD <BC
Explanation:
This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.
Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.
Let's apply these reasoning to our problem.
AB Path
this path is straight and as they indicate that the constant speed the acceleration is zero
DE path
This path is straight and since the velocity is constant the zero steps
BC path
This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression
[tex]a_{c}[/tex] = v² / r
where r is the radius of the curve and the direction of acceleration is towards the center of the curve
CD path
This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less
[tex]a_{BC}[/tex] > [tex]a_{CD}[/tex]
In summary lower accelerations are
AB = DE <CD <BC
