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an object with initial velocity v speeds up with an accelaration a ,travelling a distance L1, then it slows down with a decelaration a , and stops after travelling an additional distance L2 . if L1/L2 =k , then what is the maximum velocity of the object during its travel?

an object with initial velocity v speeds up with an accelaration a ,travelling a distance L1, then it slows down with a decelaration a , and stops after travelling an additional distance L2 . if L1/L2 =k , then what is the maximum velocity of the object during its travel?

Grade:10

1 Answers

Suraj Prajapati
36 Points
6 years ago
Let maximum velocity of the object be `v.` which would be at just before the retardation started (case I) : finding out the time taken by the object to attain that maximum velocity (v.) - v=u+at Final velocity would be" v." bcoz at this point the object begins to start retardation. v. =v+at1 Therefore t1=(v. -v) /a.............. (1)(case II) :finding the time taken by the object to come at rest from the beginning of retardation ie. initial velocity becomes "v." and finally the object stops therefore final velocity would be =0. By using the equation v=u+at 0=v. +(-a) t2 Therefore , t2 =(v.) /a....... (2) Now, we know that s=ut+1/2at²So for (case I) L1=vt1+(a/2)(t1)²On solving we get and keeping the value of t1 from eq(1)We get, L1=(v.²-v²)/2a.........(3) Similarly, we can get the L2,L2=v.²/2a..................(4) Since. L1:L2=k Therefore on keeping the value of L1 and L2 we will get from (3)... And.. (4)(v.²-v²)/v.²=k 1-(v²/v.²)=kAnd hence we get, v. =√(v²/1-k) v. =v/√(1-k) Where," v." is the maximum velocity of the object ....(before case 1)Ans. =v/√(1-k)

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