Consider two cars one a chevy, one a Ford. The Chevy is speeding along

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The discussion centers on the effects of initial velocity on stopping time and distance for two cars, a Chevy and a Ford, braking with the same constant acceleration. It clarifies that both stopping time and distance are influenced by initial velocity, as indicated by kinematic equations. Specifically, the stopping distance is particularly sensitive to initial speed, meaning higher speeds result in longer stopping distances. The conversation also highlights a common driving safety principle: faster speeds increase the risk of not stopping in time. Overall, understanding these relationships is crucial for safe driving practices.
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Consider two cars one a chevy, one a Ford. The Chevy is speeding along at 50m/s (mph) and the Ford is going half the speed at 30m/s. If the two cars brake to a stop with the same constant acceleration, are either the amount of time required to come to a stop, or the distance traveled prior to stopping influenced by their initial velocity.

I don't really understand the concept of what the question is asking. And no this is not a Homework question.
 
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Is it simply trying to parse the poorly worded paragraph? Here it is broken apart.Consider two cars one a chevy, one a Ford. The Chevy is speeding along at 50m/s (mph) and the Ford is going half the speed at 30m/s. The two cars brake to a stop with the same constant acceleration.

Is the amount of time required for each vehicle to come to a stop influenced by its initial velocity? Y/N

Prior to each vehicle stopping is the distance it traveled influenced by its initial velocity? Y/N
 


Not that important, but 30 is not half of 50.
 


It sounds to me like the question is asking you, what things depend on your speed when you undergo constant deceleration? Constant acceleration allows us to use a collection of well-worn kinematical laws, and here are two:
the change in v2 = 2 a s, where s is the distance traveled.
the change in v = a t, where t is the time.
In both cases, if you are interested in stopping, then v is just the initial velocity, and you get
s = v2/2a
t = v/a
and you clearly see that both the stopping distance, and the stopping time, depend on initial velocity v. You can also see the point commonly stressed in driver's education-- the stopping distance is especially sensitive to initial v, so if you speed, you will have a hard time stopping before you hit an object a fixed distance in front of you.
 


Are the stopping distance and/or stopping time affected by a car's initial velocity?

Kind of a long, confusing, mathematically unnecessary and inaccurate way of saying the above.
 
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