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A particle moves along a straight line OX .at a time t the distance x of the particle from O is given by x=40 +12t-t^3.how long would the particle travel before coming to rest
We have equationx=40+12t-t^3When t=0Then x=40 nowdx/dy = 12-3t^2 when velocity is zero then t=2When t=2 then x=56 now we can find how long object move before come to zero 56-40=16m
We have equation x= 40+12t-t^3. So initially take t=0 => x=40m.Now differtiate the equation with respect to time(t) which is equal to velocity (v) {v=dx/dt} now put v=0 as the particle is stopping you will get value of t. Put this value in "40+12t-t^3" and you will get final distance.on Subtracting the initial distance from this final distance we'll get x=16m.
Dear student,Please find the solution to your problem below. x = 40 + 12t – t3speed, v = dx/dt = 12 – 3t2When it comes to rest, v = 012 – 3t2 = 0Hence, t = 2 sHence, distance covered before stopping,x = 40 + 12(2) – (2)3 = 40 + 24 – 8 = 56 m Hope it helps.Thanks and regards,Kushagra
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