How close can you get to a stop sign before you start braking?

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SUMMARY

The discussion centers on calculating the stopping distance of a car before reaching a stop sign, given a maximum braking acceleration of 5 m/s². Participants emphasize the necessity of an initial velocity to solve the problem, suggesting a typical city speed of 50 km/h for calculations. The formula derived for stopping distance is x = 0.1 * v², where v is the initial velocity in m/s. The conversation highlights the importance of making reasonable assumptions in physics problems when certain data is missing.

PREREQUISITES
  • Understanding of basic kinematics, including acceleration and velocity.
  • Familiarity with the formula for stopping distance in physics.
  • Knowledge of unit conversions, particularly between km/h and m/s.
  • Ability to make reasonable assumptions in problem-solving scenarios.
NEXT STEPS
  • Study the kinematic equations for uniformly accelerated motion.
  • Learn about unit conversions between kilometers per hour and meters per second.
  • Explore real-world applications of stopping distance calculations in automotive safety.
  • Practice solving physics problems that require making assumptions based on incomplete information.
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of problem-solving techniques in real-world scenarios.

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Please help! This is urgent!

Homework Statement


1) A typical car's brakes can create a maximum acceleration of less than 5 m/s^2 . How
close can you get to a stop sign before you start braking?


Homework Equations


no idea at all since there isn't even enough information given to solve this.

The Attempt at a Solution



I don't think it's even doable, since there's no initial velocity. Or am I wrong?
 
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I'm stumped too. You need an initial velocity and a more exact acceleration than > 5m/s^2. You could do the problem with 5 m/s^2 and phrase your answer as "it needs at least x meters". But you still need an initial velocity. As your initial velocity approaches 0, your stopping distance also approaches 0. What if you were already at a stop? Then your stopping distance would be 0, regardless of how good your brakes were.

Is this a trick question and we just don't get the trick? Are you sure you worded it right?
 
I am sure I worded it right.

Actually, I just copied it straight out of the textbook.

I tried to put the Vi as 0, but that doesn't make sense. And what am I suppose to put the acceleration as? I can't really put it as <5m/s^2. :S
 
Well, here are my thoughts about this problem.
You must find a distance, measured in units of length (m). You are given acceleration, which has units of (m/s^2). Thus, you MUST have some other physical quantity, which involves units of time (s), or you won't be able to transform acceleration in such a way so as to get distance. I'd suggest you assume a typical acceleration as allowed in the city streets, such as 50 km/h and then the problem becomes fairly easy.
 
Yeah -- it could be a question to sort of those with a bit of common sense from those who read/do questions like robots.

If I came across a question in a test, the first thing I would do is write something like, "As no velocity is given, I will assume the car is traveling along a flat road at 50 km/hr."

If I was setting school tests, I'd always include a question which needed some common sense assumptions :smile:
 
Isn't it just asking you to find x in terms of vi?

No need to be sticking assumed numbers in.
 
Capuchin said:
Isn't it just asking you to find x in terms of vi?

No need to be sticking assumed numbers in.

That makes sense. Perhaps the answer is expected to be a formula. You'd still need to preface your answer with "no further than" and then use 5m/s/s as your acceleration.
 
okay given a certain speed you can calculate the stopping distance
x= [m]
v= [m/s]
a= 5 [m/s^2]
t= braking time
for a certain speed you calculate braking time:
t=v/a
x=(1/2)*a*t^2
as stated above t=v/a so -> t^2=(v/a)^2.
x=(1/2)*5*(v/5)^2
x= 2.5*(v^2)/(5^2)
x=0.1*v^2
praise me!
 

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