At the moment `t=0` a particle starts moving along the `x` axis so that its velocity projection varies as ` v_(x)=35 cos pi t cm//s`, where `t` is expressed in seconds. Find the distance that this particle covers during `t=2.80s` after the start.

Explanation

Obviously the motion law is of the from `x=a sin omegat.` and `v_(x)=omega a cos omegat. `
Comparing `v_(x)=omega cos omegat ` with `v_(x)=35 cos pi t ,` we get
`omega=pi, a =(35)/(pi),` thus`T=(2pi)/(omega)=2` and `T//4=0.5s`
Now we can write
`t=2.8s=5xx(T)/(4)+0.3(` where`(T)/(4)=0.5s)`
As `n=5` is odd, like `(4.7)` , we have to basically find the distance covered by the particlestarting from the extreme position in the time interval `0.3s`.
Thus from the Eqn.
`x=a cos omegat=(35)/(pi)cos pi(0.3)`
`(35)/(pi)-s_(1)=(35)/(pi)cos pi(0.3)` or `s_(1)=(35)/(pi){1-co 0.3 pi}`
Hence the sougth distance
`s=5xx(35)/(pi)+(35)/(pi){1-cos 0.3pi}`
`=(35)/(pi){6-cos 0.3pi}=(35)/(22)xx7(6-cos 54^(@))=60cm`