Saved Bookmarks
| 1. |
Mirror is moving towards the particle with speed 2 cm/sec. Speed of A ad B are `10sqrt(2)` and 5 cm/sec respectively in the direction shown in figure. Magnitude of velocity of image of the particle B with respect to image of A.A. `sqrt(325)cm//sec`B. `15cm//sec`C. `13cm//sec`D. `sqrt(269)cm//sec` |
|
Answer» Correct Answer - A Velocity of image of particle B `overline(V)_(B)=5(hati)+4(-hati)=hati` velocity of image of particle A `V_(A)=10(-hati)+4(-hati)-10hatj` `=14(-hati)-10hatj` Relative velocity of image `overline(V)_(BA)=overline(V)_(B)-overline(V)_(A)=hati-[14(-hati)-10hatj]=15hati+10hatj` `|overline(V)_(BA)|=sqrt(325)cm//sec` |
|