A Numerical........
Asked by abhinay.bhatta | 27th Feb, 2010, 09:10: PM
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
Regards
Team
Topperlearning.com
Answered by | 1st Mar, 2010, 04:47: PM
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