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Old Sep 30th 2018, 02:12 AM   #1
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Smile Calculating braking distance

So I need to calculate the braking distance
Relevant factors (deceleration of 5gs for a duration of 0.35seconds)
Speed = 1000km/hour

With this how can I calculate the distance it will take to brake?
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Old Sep 30th 2018, 04:20 AM   #2
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First calculate the time required. At constant deceleration -5g= -49 m/s^2, your speed will go from 1000 km/s= 1000000 m/s to 0 in 1000000/49= 20408 s.

During that time you will have gone a distance of (49/2)(20408)^2= 10204081633 m= 10204081.633 km.
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Old Sep 30th 2018, 05:58 AM   #3
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Originally Posted by HallsofIvy View Post
First calculate the time required. At constant deceleration -5g= -49 m/s^2, your speed will go from 1000 km/s= 1000000 m/s to 0 in 1000000/49= 20408 s.

During that time you will have gone a distance of (49/2)(20408)^2= 10204081633 m= 10204081.633 km.
That's some car Ivy !! a million meters a second is 2 Million MPH

Initial speed 1000 Km per 3600 secs = 277m/sec ....deceleration 49m/s2 brakes applied only for 0.35sec v=u + ft2

v = 277- 49x .35 x .35 ..................271m/sec this is the speed after breaking

You want to know "the distance it will take to brake" ... the car will still be moving after this ....

Average velocity during breaking is 274m/sec ...in 0.35 sec it will travel 96 meters

Distance travelled during breaking is 96 meters

Last edited by oz93666; Sep 30th 2018 at 06:38 AM.
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Old Sep 30th 2018, 11:33 AM   #4
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Smile What about this approach?

Using s = vt +1/2at^2
v = 277.778 ms (1000km/hr -> m/s)
a = 5*9.81
t = v/a (277.778/5*9.81) =5.6632
so
s = 2359meters
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Old Sep 30th 2018, 10:34 PM   #5
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Originally Posted by bcornelis View Post
Using s = vt +1/2at^2
v = 277.778 ms (1000km/hr -> m/s)
a = 5*9.81
t = v/a (277.778/5*9.81) =5.6632
so
s = 2359meters
Error in line 4

You have already specified breaks were applied for 0.35 secs ...

How far will it travel in 0.35 secs when max speed is 277 ???
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braking, calculating, distance



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