Physics Problem involving Kinematics

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Homework Help Overview

The discussion revolves around a kinematics problem involving a car's braking distance and the maximum deceleration it can achieve. Participants explore how to express the braking distance in relation to initial velocity and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the feasibility of calculating stopping distance without a defined initial velocity, with some suggesting the use of variables to express the answer. Others mention dimensional analysis as a potential tool for understanding the relationship between variables.

Discussion Status

The conversation is active, with participants providing guidance on expressing the solution in terms of variables. There is an exploration of relevant kinematic equations, but no consensus has been reached on a specific method or final answer.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the lack of an initial velocity and how that affects the formulation of the braking distance. The discussion reflects the challenge of working with unknowns in a physics context.

bbface_assassin
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1) A typical car's brakes can create a maximum acceleration of less than 5 m/s2 . How
close can you get to a stop sign before you start braking?

I don't think it's even doable, since there's no initial velocity. Or am I wrong?
 
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you are correct, sir. though, you could write the answer in terms of the unknown v_0.
 
Oh. So then my answer would be in variables, and not actual numbers?
 
yep. you can write the answer for the shortest braking distance in terms of the initial velocity and the acceleration... you might be able to guess it (up to a factor of 1/2) from dimensional analysis (i.e., looking at the units).
 
So can I use V^2 = Vo^2 + 2ax as the formula for solving this?
 
yeah, since you know what v has to be when the car has stopped. you can solve for x in terms of v0 and a.
 

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