Saved Bookmarks
| 1. |
A man of mass 70 kg is standing at one end of a stationary, floating barge of mass 210 kg. he then walks to the other end of the barge, a distance of 90 meters. Ignore any frictional effect between the barge and the water. Q. If the man walks at an average velocity of 8m/s what is the average velocity of the barge? |
|
Answer» Solution :Let he time it takes the man to walk ACROSS the barge be denoted by r, then `t=(90m)/(8m//s)`. In this AMOUNT of time, the barge moves a DISTANCE of 22.5 meters in the OPPOSITE, so the velocity of the barge is `v_("barge")=(-22.5m)/(t)=(-22.5m)/((90m)/(8m//s))=-(180m^(2)//s)/(90)=-2m//s` |
|