CBSE Class 11-science Answered
1?
Asked by smanishkumar2002 | 02 Jun, 2018, 07:58: AM
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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