Calculations in regards to a car coming to a stop, help is needed.

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The discussion revolves around calculating stopping distances for a car in a school zone, emphasizing the importance of adhering to a speed limit of 40 km/h instead of 50 km/h. The initial formula used was incorrect due to unit conversion errors and misplacement of the friction coefficient in the equation. After correcting the initial velocity to meters per second, the user recalculated and found a stopping distance of approximately 9 meters, which aligns better with expected results. The conversation highlights the significance of proper unit conversion and formula application in physics calculations. Overall, the user received guidance on correcting their calculations to better understand stopping distances.
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I have been given a physics assignment which requires that I go over the physics of why 40km/h is a better speed to be followed in a school zone rather than 50km/h. (must include calculations for stopping distances etc.)

I used the formula Vo ^ 2 / 2 x μ x g.
It should give me the stopping distance if I am correct, Vo being the inital velocity, μ the friction coefficient and g for gravity. I first tried using 40km/h for Vo, .7 for μ and 9.8 for g.

40^2/2x.7x9.8

I use this and end up with 116.61...kilometres, this is evidently wrong and I am not sure what I am doing wrong as it should come out to 20-30 metres.
Can somebody please help?
Thanks.
(P.S I am in grade 11, so go easy on me :frown:)
 
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You forgot to convert km/hr into m/s. It is essential that units are consistent.
 
Oh that gives me an answer of 13.72, I feel stupid now. Thanks for the help!
 
connormaphone said:
Oh that gives me an answer of 13.72, I feel stupid now. Thanks for the help!
13.72 m? Show how you arrived at that answer.
 
40000^2/2x.7x9.8
Just realized this is wrong as well. I don't even know.
 
Actualy, just made it 11.1 m/s which does equal 40km/h, I put it into the equation and ended up with approximately 9 metres.
 
connormaphone said:
Actualy, just made it 11.1 m/s which does equal 40km/h, I put it into the equation and ended up with approximately 9 metres.
Now, you're talking! Looks Good!:smile:
And welcome to PF!
 
connormaphone said:
I have been given a physics assignment which requires that I go over the physics of why 40km/h is a better speed to be followed in a school zone rather than 50km/h. (must include calculations for stopping distances etc.)

I used the formula Vo ^ 2 / 2 x μ x g.
It should give me the stopping distance if I am correct, Vo being the inital velocity, μ the friction coefficient and g for gravity. I first tried using 40km/h for Vo, .7 for μ and 9.8 for g.

40^2/2x.7x9.8

I use this and end up with 116.61...kilometres, this is evidently wrong and I am not sure what I am doing wrong as it should come out to 20-30 metres.
Can somebody please help?
Thanks.
(P.S I am in grade 11, so go easy on me :frown:)

Your starting formula is incorrect. The μg should be in the denominator. That is the deceleration. The larger the deceleration, the shorter the stopping distance.

Chet
 
It is in the denominator. Parentheses would help, but I think it is in proper PEMDAS, per my recollect.
 
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Got it, thanks for the help guys!
 
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