contestada

A car is stopped at an entrance ramp to a freeway; its driver is preparing to merge. At a certain moment while stopped, this driver observes a platoon of vehicles a distance x0 upstream and initiates the merge maneuver. The platoon approaches the entrance ramp at a constant speed v. The stopped car can accelerate from speed 0 to v at uniform acceleration rate a .

Find the latest time at which the stopped car can safely start the merge maneuver; i.e., derive an expression for the "latest safe start time" in terms of the variables x0, v and a. Assume this latest time enables the platoon to maintain its constant speed and "just touch" the merging car's trajectory. Ignore the physical dimensions of the car.

Respuesta :

emaduet2012

Answer:

T = [tex]\sqrt{\frac{2X_{0} }{a} }[/tex]

Explanation:

Given,

Initial Velocity, u=0

As platoon is moving with constant velocity v,

Final Velocity, v=v

The vehicle starts from 0 to v at constant acceleration of a,

Relevent expressions:

v=u+at...........................(1)

v2=u2+2as.....................(2)  

[tex]V^{2}[/tex] = 2aS, as S = [tex]X_{0}[/tex],

[tex]V^{2}[/tex] = 2a[tex]X_{0}[/tex],

From(1)

v=at

Hence

(at)^2=2a[tex]X_{0}[/tex]

[tex]T^{2}[/tex]   =  [tex]\frac{2aX_0}{a^{2} }[/tex]

T = [tex]\sqrt{\frac{2X_{0} }{a} }[/tex]

This is the final expression for time.

Otras preguntas