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ICSE Class 9 Answered

ques no-8
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Asked by baajirao106 | 04 Oct, 2018, 06:41: PM
answered-by-expert Expert Answer
Acceleration of Car-A  begin mathsize 12px style equals space fraction numerator c h a n g e space i n space s p e e d over denominator t i m e end fraction space equals fraction numerator 30 minus 0 over denominator 40 end fraction equals 0.75 space m divided by s end style
distance travelled by car-A in 40s :- (1/2)×a×t2 = (1/2)×0.75×40×40 = 600 m
distance travelled by Car-B in 40 s = 20×40 = 800 m
 
at 40s, distance between the cars = 800-600 = 200 m
 
after 40s both the cars are travelling with constant speed and distance between them is 200 m.
relative velocity = 30-20 = 10 m/s
hence time taken to reduce the gap distance = 200/10 = 20 s
 
If we add the initial time 40s, then car-A will cross car-B at 40+20 = 60 s
Answered by Thiyagarajan K | 04 Oct, 2018, 07:42: PM

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