The speed of a body moving on a straight trach varies according to `v=(4)/(5)t^2+3` for `0lttlt5` sec and `v=3t+8` for `tgt5` sec. If dinstances are in meters, find the distance moved by a particle at the end of 10 sec.

Answer

`((145)/(3)+(305)/(2))`m

Explanation

`(ds)/(dt)=(4)/(5)t^2+3`
`impliesS_1=int_0^(S_1)ds=int_0^5((4)/(5)t^2+3)dt`
`S_1=[(4)/(15)t^3+3t]_0^5=(145)/(3)m`
`S_2=int_0^(S_2)ds=int_5^10(3t+8)dt`
`S_2=[(3t^2)/(2)+8t]_5^10=(305)/(2)m`
`S=S_1+S_2=((145)/(3)+(305)/(2))m`