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A ship of length l-150m moving with velocity v_(s)=36 km h^(-1) on the sea, suddenly discovered straight head a siking boat people having met an acceleident. A rescue boat has been lowered from the mid of the ship, which went to the sinking boat with speed v_(b)=72 km h^(-1). When the rescue boat was x_(0)=3.0 km away, The rescue boat reaches the sinking boat spends t_(0)=1.0mim there to take the people on board, and then retuned with the same speed to the time taken in the whole rescue it was lowerd. Derermine the time taken in the whole rescue operation from the moment the rescue boat was lowerd to the moment therescue boat returned to the ship. |
Answer» Solution :Time taken by rescue boat to teach from `C` to`E` . `t_(10=(75))/(v_(B)-v_(s)) =(75)/(20-10) =7.5s` Time taken by rescue boat to GO from `E to `S`: `t_(2) =(300)/(v_(B)) =(3000)/(20) =150 s` Time spent `t_(0)=1 min =60 s Distance travelled by ship time, `t_(2)+t_(0)=v_(s)(15+60) =10 xx 210 =2100 m` Time taken by rescue boat to REACH from `S` to `C`, `t_(3)=((3000-2100)+75)/(v_(B)+v_(s)) =(975)/(30) =32.5 s` Total time : `T=t_(1)+t_(2)+t_(0)+t_(3)`. |
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