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^(-2)`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Options

A. 0.68 m `s^(-2)`
B. 0.86 m `s^(-2)`
C. 0.56 m `s^(-2)`
D. 0.76 m `s^(-2)`

Answer

B

Explanation

Here, `v=27kmh^(-1)=27xx(5)/(18)ms^(-1)`
`v=(15)/(2)ms^(-1)=7.5ms^(-1),r=80m`
Centripetal acceleration, `a_(c)=(v^(2))/(r)`
`a_(c)=((7.5ms^(-1))^(2))/(80m)~~0.7ms^(-2)`
Tangential acceleration `a_(t)=0.5ms^(-2)`
Magnitude of the net acceleration is
`a=sqrt((a_(c))^(2)+(a_(t))^(2))=sqrt((0.7)^(2)+(0.5)^(2))~~0.86ms^(-2)`