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. `54.44^@`
B. `64.44^@`
C. `74.44^@`
D. None of these

Answer

A

Explanation

Speed of cyclist , `27kmh ^(-1)=27xx(5)/(18)ms^(-1)=(15)/(2)ms^(-1)`
Radius of circular turn , r 80m.
Centripetal acceleration `a_c`is given by
`a_c=(v^2)/(r)=(15/(2))^2/(80)=0.703ms^(-2)`
Let P be the point at which the cyclist applies brakes , then the tangentialacceleration `a_t` (which will be negative ) will act opposite to velocity of cyclist and is given by
`a_t=("change in velocity ")/(Time)=0.5ms^( -2)`
Let `theta `be the angle made by the net acceleration with the velocity of cyclist , then
`tan theta=(a_c)/(a_t)=(0.703)/(0.5)=1.406`
`therefore " "theta=54.44^@`