Stopping distance

  • Thread starter Travian
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  • #1
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Homework Statement


The maximum straight-line deceleration of a racing car under braking is 5 m s-2. What is the minimum stopping distance of the car from a velocity of 108 km h-1?



Homework Equations



½ V x t

The Attempt at a Solution



V = 108km/h=30m/s

t = 30/ 5 = 6 m s-2

½ V x t = ½ x 30 x 6 = 90

Answer: Distance = 90 m

___________

I found this formula ½ V x t (to determine stopping distance) on the internet, but im not sure if its correct.

Are there any mistakes?

Thank You
 

Answers and Replies

  • #2
28
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It seems like your formula for distance is the average distance travelled in a given time for an object moving at constant speed. I would Instead try the formula:
vf 2 = vo 2 + 2 a [tex]\Delta[/tex] x

Note that you are decelerating so the plus in front of the 2 is actually a minus here.
 
Last edited:
  • #3
64
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so: vf = 30 m/s -2 x 5 m/s...
what is this delta x? sorry, haven't slept this night at all, feeling a bit dizzy now:/
 
  • #4
28
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Sorry, delta x here stands for the change in position of the car.

Basically, how far it goes in stopping.

Edit: Sorry I'm without a calculator so I hadn't noticed but this gives the same answer you already had.
 
  • #5
64
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90 meters?
 
  • #6
28
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Yes, 90 meters. Aside from the units in your 't', everything seems to be in order.
 
  • #7
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Thank you.

By the way, whats wrong with t?
 
  • #8
28
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Units: I'm a little mixed up by your scheme but it looks like you have t (time) in meters per second per second, instead of just seconds.
 
  • #9
64
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Oh seriously... I fixed that. Thanks again:)
 

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