A cyclist is riding with a speed of `27 km h^(-1)`. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate `0.5 ms^(-1)`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Explanation

Here v=`27 km//h = 27 xx 5/18 m//s=7.5m//s,r=80m`
and tangential acceleration `a_(t)=-0.50m//s^(2)`.
`therefore` Centripetal acceleration `a_(c)= v^(2)/r = (7.5)^(2)/(80) = 0.70 ms^(-2)` (radially inwards).
Thus, as shown in fig. above, two acceleration are acting in mutually perpendicular directions. If `veca` be the resultant accelerations, then
`|veca|=sqrt(a_(1)^(2)+a_(c)^(2)) = sqrt((0.5)^(2)+(0.7)^(2))= 0.86ms^(-2)`

and `tanbeta= a_(c)/a_(t) = 0.7/0.5 = 1.4`
`rArr beta=tan^(-1)(1.$ = 54.5^(@)` from the direction of negative of the velocity.