Question

A cyclist is riding with a speed of 27 km h-? As he
approaches a circular turn on the road of radius
80 m, he applies brakes and reduces his speed at the
constant rate of 0.50 m sl every second. What is the
magnitude of the net acceleration of the cyclist on the
circular turn?​

Answer 1

Answer:

1/4{√(15√5+20)}

Explanation:

There will be two accelerations, one at centre & another one backwards.

A1 = v^2/r = [(15/2)^2 ]/r

(since 27k/h = 27*5/18m/s)

A2 = -1/2 m/s^2

Net acceleration = √[(A1)^2+(A2)^2 ]