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CBSE Class 11-science Answered

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Asked by smanishkumar2002 | 02 Jun, 2018, 07:58: AM
answered-by-expert Expert Answer
we need to use two equations of motion (1) v2 = u2+2aS and (2) S = ut+(1/2)at2
 
where v and u are final and initial speed respectively. a is acceleration and S is distance travelled in t seconds
 
with the given information eqn(1) and (2) are wriiten as
 
225 = u2+120a .....................................(3)
 
60 = 6u+18a .............................(4)
 
(4) is re written as ,  10 = u+3a ; hence a =(1/3)(10-u)  ...........(5)
 
By substituting for acceleration from eqn.(5) in (3)  we get
 
u2 - 40u +175 =0 ;
 
(u-5)(u-35) = 0;  u can not be equal to 35, hence u = 5 m/s.
 
acceleration = (1/3)(10-5) = 5/3 m/s2
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