Saved Bookmarks
| 1. |
The displacement (in metre) of a particle moving along x-axis is given by `x=18t +5t^(2).` Calculate (i) the instantaneous velocity `t=2 s` (ii) average velocity between `t=2 s` to `t=3 s` (iii) instantaneous acceleration. |
|
Answer» Give , `x=18t +5t^(20` (i) Velcity, `v=(dx)/(dt) =d/(dt) (18 t+ 5t^(2) ) =18 + 10 t` At `t=2 s`, the instantaneous velcoty is `v_(i) =18 + 10 xx 2 = 38 ms^(-1)` (ii0 Displacement at `t_(1)=2 s` is. `x_(1) =18 xx 2 xx 5 xx 2^(2) =56 m` Displacement at `t_(2) =3 s` is, `x_(2) =18 xx 3+5xx 3^(2) =99 m` Average velocity, `v_(av) =(x_(2)-x_(1))/(t_(2)-t_(1)) =(99- 56)/(3-2) =43 ms^(-1)` (iii) Acceleration, `a=9dv)/(dt) =d/(dt) (18 +10t) =10 ms^(-2)`. |
|