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an express train moving at 30m/s reduses its speed to 10m/s in a distance of 240m if the breaking force is increased by 12.5% in the begining find the distane it that travels before coming to rest
3 years ago

Answers : (1)

deeksh bharath
11 Points
							
given that the train is moving initially at 30m/s and then decelerates to 10m/s then
v2_u2=2as
(10)2-(30)2=2(-a)(240)
100-900=(-480a)
-800= -480a
a= -5/3m/s2
again tis given that if initially the breaking force is 12.5%
the by f=ma, as f increases ,aceleration also increases,therefore
new aceleration is a( 1+12.5/100 )(-50/3)
a=1.875 or 15/8m/s2
now use the sformula u2/2a
             d = (30)2/2*(-15/8)m
               d = 240m
therefore the train will come to rest or stom at the ditance of 240m if the breaking force is 12.5% in the initial.
 
9 months ago
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