The distance travelled by a particle moving along a st. line is given by `x=4 t + 5t^(2) +6^(3)` mette.
Find (i) the initial velcity of the particle (ii) the velcoty at the end of `4 s` and (iii) the acceleration of the particle at the end of `5 second.

Explanation

(i) velocity,
`v=(dx)/(dt) =d/(dt) (4t +5t^(2) +6t^(3))`
`=4 +10 t + 18 (0) ^(2) =4 m// s`
(ii) Velocity of the particle when `t=4 s`
`v=4 +40 xx 4 + xx (4)^(2) =332 m//s`
`v=4 + 10 xx 4 18 xx (40 ^(2) =332 m//s`
(iii) Accelration,
`a=9dt)/(dt) d/(dt) (4+10 t+18)
`= 0 + 10 + 36 t =10 36 t`
When `t=5 s`, a=10 + 36 xx 5=190 ms^(-2)` .