A particle moves along the `x` axis according to the law` x=a cos omega t`. Find the distance that the particle covers during the time interval from `t=0` to `t`.

Explanation

From the motion law , `x=a cos omegat`,, it is obvious that the time taen to cover the distance equal to the amplitude `(a)`, starting from extreme position equals `T//4`.
Now one can write
`t=n(T)/(4)+t_(0)` (where`t_(0) lt (T)/(4)` and `n=0,1,2,.....)`
As the particle moves according to the law, `x=a cos omegat` ,
So at `n=1,3,4 ......` or for odd `n` values it passes through the mean position and for even number of `n` it comes to an extreme position ( if `t_(0)=0)`
Case `(1)` when `n` is an odd number `:`
In this case, from the equation
`x=+- a sin omegat`, , if the `t` is counted from `nT//4` and the distance covered in the time interval to becomes, `s_(1)=a sin omega t_(0)=a sin omega (t-n(T)/(4))=a sin (omegat-(npi)/(2))`
Thus the sought distance covered for odd `n` is
`s=na +s_(1)=na+a sin (omegat-(n pi)/(2))=a[n+sin (omegat-(npi)/(2))]`
Case `(2)` , when `n` is even, In this case from the equation
`x=a cos omegat`, the distance covered `(s_(2))` in the interval `t_(0)` is given by
`a-s_(2)=a cos omegat _(0)=a cos omega(t-n(T)/(4))=a cos (omegat-m(pi)/(2))`
or , `s_(2)=a[1-cos(omegat-(npi)/(2))]`
hence the sought distance for `n` is even
`s=na+s_(2)=na+a[1-cos(omegat-(npi)/(2))]=a[n+1-cos(omega-(npi)/(2))]`
In general
`s={(a[n+1-cos (omegat-(mpi)/2)],n is even),(a[n+sin(omegat-(npi)/(2))],n is odd):}`