A cyclist is riding with a speed of 27kmh^(-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 0.5ms^(-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 27kmh^(-1). As he approaches a cir

1.	A cyclist is riding with a speed of 27kmh^(-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 0.5ms^(-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)=0.5ms^(-2)` `a=sqrt(a^(2)-a^(2))=0.86ms^(-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.`