A Numerical........

Asked by abhinay.bhatta | 27th Feb, 2010, 09:10: PM

Expert Answer:

Dear Student

Relative velocity of  C with respect to A= 54+36= 90 kmph

Relative velocity of  B with respect to A= 54-36= 18 kmph

Time required for C to reach A (1KM)= 1/90    hr

If  'B'  decides to overtake  'A'  before  'C'  does, B should reach A in a time <1/90 hr

Consider B

Time=1/90 hr

Distance=s=1km

Initial velocity= u=18kmph

Final Velocity= v

Acceleration   a= v-u   /t

s= ut +  1/2   a  t^2

   = ut     +    1/2   X  v-u   /t      Xt^2  =  ut     +    1/2   X ( v-u)t    =t (u+0.5(v-u))

==> 1=1/90    X  (  18 + 0.5Xv- 0.5X18  ) 

(  18 + 0.5Xv- 0.5X18  )    =90

V=162 kmph

a= (162-18)/ (1/90)  = 12960 km/hr^2= 1 m/sec^2

So acceleration must be > 1 m/sec^2

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Answered by  | 1st Mar, 2010, 04:47: PM

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