Saved Bookmarks
| 1. |
Derive the equations of uniformaly accelerated motion by graphical method. |
Answer» Solution :Now at `t=0` the velocity of the particle be `v_(0)` and at t =t the velocity of the particle be v.  `therefore` Acceleration a = SLOP of line AB `= (v-v _(0))/( t -0)` `therefore a = (v - v _(0))/(t )` `therefore v = v _(0) + at ""...(1)` Now displacement of a particle during time t is equal to the area of region OABCD in `v to t` graph. `therefore x =` Area of RECTANGLE OACD + Area of `Delta ACB` `=v_(0) t + 1/2 (v - v _(0)) t` `therefore x = v _(0) t + 1/2 at ^(2) ""...(2)` but AVERAGE velocity, ` x = ((v + v _(0))/( 2 )) t ""...(3)` For, EQUATION (3) and equatin (1), `t = (v-v_(0))/(a)` `therefore x = ((v + v _(0))/(z )) ((v - v _(0))/(a ))` `therefore v ^(2) -v _(0) ^(2) = 2ax` Above equation can be written as follows : `v = v _(0) + at` `x =x _(0) +v_(0) t + 1/2 at ^(2)` `2a (x -x_(0)) =v^(2) -v_(0) ^(2)` |
|