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if the velocity v of a paticle moving along a straight line decreases linearly with its displacement from 20m/s to a value apporoaching zero at s=30m,determine the acceleration of the particle when s=15m.
consider attatched graph... the relation b/w v & x comes to be v = -20/30 * x + 20 now differentiate with respect to time(t) we get dv/dt = (-20/30)*(dx/dt) as dv/dt = a & dx/dt= v put the respective values in the eqn => a = (-20/30)v => a = (-20/30)*(-20/30 * x + 20) now to find a at x= 15 put x= 15 in above eqn. we get : a = -20/3
consider attatched graph...
the relation b/w v & x comes to be
v = -20/30 * x + 20
now differentiate with respect to time(t)
we get dv/dt = (-20/30)*(dx/dt)
as dv/dt = a & dx/dt= v
put the respective values in the eqn
=> a = (-20/30)v
=> a = (-20/30)*(-20/30 * x + 20)
now to find a at x= 15
put x= 15 in above eqn. we get : a = -20/3
Given : Decreasing velocity is = -20 m/s Its value approaches to zero i.e. final velocity = 0 m/s Distance travelled = 30 m Acceleration a = v2 - u2 / 2s = 0-400 / 2*30 = -400/60 =-40/6 = -6.6 m/s2 At s=15m a = 6.6m/s2 only as velocity while decelerating changes with constant rate by which velocity remains constant. HOPE MY ANSWER IS RIGHT & YOU WOULD GIVE ME A POINT
Given : Decreasing velocity is = -20 m/s
Its value approaches to zero i.e. final velocity = 0 m/s
Distance travelled = 30 m
Acceleration a = v2 - u2 / 2s
= 0-400 / 2*30
= -400/60
=-40/6
= -6.6 m/s2
At s=15m a = 6.6m/s2 only as velocity while decelerating changes with constant rate by which velocity remains constant.
HOPE MY ANSWER IS RIGHT & YOU WOULD GIVE ME A POINT
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