A BULLET STOPS AFTER TRAVELLING A DISTANCE OF XM UNDER THE ACTION OF A UNIFORM RETARDING FORCE .HOW FAR WILL IT GO BEFORE BEING STOPPED BY THE SAME FORCE IF ITS VELOCITY IS DOUBLED?
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v^2=2ax but a=-v/t here on doubling very well get 4v^2=4v/tx. so it would travel same distance
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Stopping distance(s) is given by:
s = (u^2)/2a.
(using v^2-u^2=2as . and putting v=0)
where u = initial velocity and a= retardation.
it is given that:
X= (u^2)/2a ..........(1)
when its velocity doubled (=2u):
X'=[(2u)^2]/2a............(2)
where X' is the stopping distance in later case. as force is uniform in both case 'a ' does not change.
X'=4×[ (u^2)/2a ]
then using (1) and (2) we get,
=> X'=4X.
it will go just 4 times more far than in former case.
s = (u^2)/2a.
(using v^2-u^2=2as . and putting v=0)
where u = initial velocity and a= retardation.
it is given that:
X= (u^2)/2a ..........(1)
when its velocity doubled (=2u):
X'=[(2u)^2]/2a............(2)
where X' is the stopping distance in later case. as force is uniform in both case 'a ' does not change.
X'=4×[ (u^2)/2a ]
then using (1) and (2) we get,
=> X'=4X.
it will go just 4 times more far than in former case.
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