A cyclist is riding with a speed of 27 kmh^(-1) . As he approaches a circular turn on the road of radius 30 m, he applies brakes and reduces his speed at the constant rate 0.5 ms^(-2). What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

A cyclist is riding with a speed of 27 kmh^(-1) . As he approaches a c

1.	A cyclist is riding with a speed of 27 kmh^(-1) . As he approaches a circular turn on the road of radius 30 m, he applies brakes and reduces his speed at the constant rate 0.5 ms^(-2). What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
Answer» Solution :`a_(c)=(v^(2))/(r)=0.7ms^(-2)` `a_(T)=0.5 ms^(-2)` `a=sqrt(2_(C)^(2)+a_(T)^(2))=0.86 ms^(-2)` If `theta` is the angle between the NET acceleration and the VELOCITY of the cyclist, then `theta=tan^(-1)((a_(c))/(a_(T)))=tan^(-1)(1.4)=54^(@)28'`