HELP -How many meters does it take for a car to stop?

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To determine how many meters it takes for a car to stop from 45 m/s with a constant deceleration of -11 m/s², the kinematic equation can be used, specifically the one that relates initial velocity, final velocity, acceleration, and distance. The equation is: distance = (final velocity² - initial velocity²) / (2 * acceleration). For a car traveling at 45 m/s, the stopping distance calculates to approximately 103.125 meters. If the car were traveling at half that speed (22.5 m/s), the stopping distance would be significantly less, calculated using the same formula. Understanding these kinematic equations is essential for solving such motion problems effectively.
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A race car can be slowed with a constant acceleration of -11 m/s squared
If the car is going 45 m/s, how many meters will it take to stop?
How many meters will it take to stop a car going half as fast?
 
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Look up the kinematic equations of motion
 
1. what are those
2. i don't know how that helps.. all i need is an answer
 
merceb49 said:
1. what are those

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html

merceb49 said:
2. .. all i need is an answer

PF doesn't work like that. We will help you arrive at the answer.
 
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ok well i do not know what type of formula, how do i get it?
 
merceb49 said:
ok well i do not know what type of formula, how do i get it?

Read the link I posted and see which variables you have in your problem.
 
thank you for the link, but none of those equations work, there is no time involved in the situation. Are there any other ways, I'm looking for how many meters it will take for the car going 45m/s to stop at the constant rate of -11m/s squared
anything else, please?
 
merceb49 said:
thank you for the link, but none of those equations work, there is no time involved in the situation.
Your observation that you do not need time to solve the problem is correct. The observation that all those equations involve time is not- look carefully.
 
thank you so much.
 
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