InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A particle starts to move in a straight line from a point with velocity 10 m s^(-1) and acceleration - 2.0 m s^(-2). Find the position and velocity of the particle at (i) t = 5 s, (ii) t' = 10 s. |
|
Answer» Solution :Given u =10 m `s^(-1) ` a = 2.0 `m s^(-1)` (i) Displacement at t = 5 s is `S = ut +(1)/(2) a t^(2)` `= 10 XX5 +(1)/(2) xx(-2.0) xx(5)^(2)` = 50 -25 =25 m i.e., after 5 s the PARTICLE will be at distance 25 m from the starting point. VELOCITY at t = 5 s is v = u +a t or v = 10 + (-2.0) `xx` 5=0 i.e., the particle is momentarily at rest at t = 5 s. (ii) Diplacement at t = 10 s is `S= ut +(1)/(2) a t^(2)` `= 10 xx10 +(1)/(2) xx(-2.0) xx(10)^(2)` =100- 100 =0 (zero ) i.e., after 10 s the particle has come BACK to the starting point Velocity at t = 10 s is v = u + at or v = 10 + `(-2.0 ) xx 10 = -10 m s^(-1)`. i.,e., velocity is 10 `m s^(-1)` towards the starting point (i.e., opposite to the initial direction of motion ). |
|