1.

A particle moves in a circular path such that its speed `1v` varies with distance `s` as `v = sqrt(s)`, where `prop` is a positive constant. Find the acceleration of the particle after traversing a distance `s`.

Answer» Total acceleration, `a = sqrt(a_t^2 + a_r^2) = sqrt(((d v)/(d t))^2 + ((v^2)/(R))^2)`
where `v = prop sqrt(s)`
Differentiating `v = prop sqrt(s)` with respect to time, we have
`( dv)/(d t)= prop (s^(-1//2))/(2) (d s)/(d t)=`
Substituting `(d s)/(d t) = prop sqrt(s)`, we have `(d v)/(d t) = (prop^2)/(2)`
Now substituting `dv//dt` and `v` in the expression of `a`, we have
`a = sqrt(((prop^2)/(2))^2 + [(prop sqrt(s))/(R)]^2`
This gives `a = prop^2 sqrt((1)/(4) + (s^2)/(R^2))`.


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