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. `0.68ms^(-2)`
B. `0.86ms^(-2)`
C. `0.56ms^(-2)`
D. `0.76ms^(-2)`

Answer

B

Explanation

Here, `v= kmh^(-1)27xx(5)/(18) ms^(-1)=7.5 ms^(-1), r=0 m` Centripetal acceleration, `a_(c)=(v^(2))/(r=((7.5 ms^(-1))^(2))/(80 m)=0.7 ms^(-2)`
Tangential acceleration `a_(t)=0.5ms^(-2)`
`:.` agnitude of net acceleration is `a=sqrt((a_(c))^(2)+(a_(t))^(2))=sqrt((0.7)^(2)+(0.5)^(2))=0.86 ms^(-2)`