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A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2, how l...Asked by Rayan
A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2, how long does it take to catch the speeder?
Answers
Answered by
bobpursley
they travel the same distance, in the same time
change 100km/hr to 27.8m/s
ds=27.8*t
dp=1/2 3.6*t^2
set them equal
27.8 t= 1.8 t^2 solve for t.
change 100km/hr to 27.8m/s
ds=27.8*t
dp=1/2 3.6*t^2
set them equal
27.8 t= 1.8 t^2 solve for t.
Answered by
Rayan
why are you multiplying by 1/2
Answered by
Damon
with constant acceleration a:
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
here Vi = Xi = 0
so
x = (1/2)at^2
v = Vi + a t
x = Xi + Vi t + (1/2) a t^2
here Vi = Xi = 0
so
x = (1/2)at^2
Answered by
Sean
how do you find the distance it has to travel before it catches speeder?
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