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 of `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 ms^(-2)`
B. `0.86 ms^(-2)`
C. `0.56 ms^(-2)`
D. `0.76 ms^(-2)`

Answer

b

Explanation

Here, `v=27kmh^(-1)=27xx5/18ms^(-1)`
`v=15/2=7.5ms^(-1)`
`r=80m`
Centripetal acceleration, `a_(c)=(v^(2))/r`
`a_(c)=((7.5ms^(-1))^(2))/(80m)=0.7 ms^(-2)`
Tangential acceleration, `at=0.5 ms^(-2)`
Magnitude of the net acceleration is
`a=sqrt((a_(c))^(2)+(a_(t))^(2))=sqrt((0.7)^(2)+(0.5)^(2))=0.86 ms^(-2)`