Saved Bookmarks
| 1. |
A particle starts from origin af i = 0 with a velocity 5.0 hati m//s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0i+2.0j) m//s^(2) ?. (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 mn? (b) what is the speed of the particle at this time? |
|
Answer» Solution :The position of the PARTICLE is given by `r(t)=v_(0)t+(1)/(2) "at"^(2)=5.0hati t+(1//2)(3.0 hati+2.0 hatj)t^(2)` `=(5.0t+1.5t^(2))hati+1.0 t^(2) hatj` THEREFORE `x(t)=5.0t+1.5t^(2), y(t)=+1.0 t^(2)` Given `x(t)=84 m , t= ` `5.0t+1.5t^(2)=84 rArr t=6s` At `t=s, y=1.0 (6)^(2)=36.0 m` Now the velocity `V=(dr)/(dt)=(5.0+3.0t)hati+2.0 t hatj` At `t=6 s, v=23. hatj+12.0 hatj` Speed `=|v|= sqrt(23^(2)+12^(2))=26 m s^(-1)` |
|