1.

If the displacement x of a particle (in metre) is related with time (in second) according to relation `x = 2 t^3 - 3t^2 +2 t + 2` find the position, velocity and acceleration of a particle at the end of 2 seconds.

Answer» Correct Answer - `10 m ; 14 m//s ; 18 m//s^2`
`x = 2t^3 - 3t^2 +2 t +2`
velocity `=(dx)/(dt) = 6t^2 - 6t +2. "Acceleration" =(d)/(dT)((dx)/(dt)) = 12t -6`
When `t = 2 sec, x =2xx2^3 - 3xx2^2 +2xx2+2 =16 -12 +4 +2 =10m`
Velocity ` =6xx2^2 - 6xx2+2 =14m//s,` Acceleration `=12xx2 -6 =18m//s^2`


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