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Mathematics 19 Online
OpenStudy (anonymous):

In a driving manual, it is suggested that, when driving at 13m/s (about 45 km/hr), a driver should always keep a minimum of two car lengths between the driver's car and the one in front.

OpenStudy (anonymous):

Suggest a scientific justification for this safety tip.

OpenStudy (anonymous):

@UnkleRhaukus @experimentX Can you help?

OpenStudy (experimentx):

what would happen if there were less distance and the front car would suddenly stop?|dw:1349168122867:dw|

OpenStudy (anonymous):

to get enough time to stop ur car ... if the one infront ot u suddenly stops..

OpenStudy (anonymous):

but why driving exactly at 13m/s?

OpenStudy (anonymous):

and 2 car lengths is too much to stop the car.

OpenStudy (experimentx):

try to work out ... final speed = 0 initial speed = 13 m/s deceleration = ?? distance = ?? find the relation between distance and de-acceleration. note that too much deacceleration will kill you.

OpenStudy (anonymous):

length of two cars = 10m. (approx) u = 13m/s by using equation of uniformed motion , t = 1.5s. But whats the relation between the given numerical quantities? 13m/s and distance of 2 cars?

OpenStudy (experimentx):

v^2 = u^2 + 2as 13^2 = 2*a*s let 30 m/s^2 kills you ... what should be the distance.

OpenStudy (unklerhaukus):

s=d/t t=d/s say your car is 5 m and you are traveling at 13 m/s two car lengths will be 10 m you will have 10/13 ~ 3/4 seconds of time to react

OpenStudy (anonymous):

@experimentX around 2.81m

OpenStudy (experimentx):

yeah you would end up dead at 2.81 ms huge deacceleration means ... when you stop suddenly, |dw:1349169386925:dw|

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