Question

A cyclist is riding with a speed of `18kmk^(-1)`. As he approaches a circular turn on the road of radius `25sqrt2m`, he applies brakes and reduces his speed at the constant rate of `0.5ms^(-1)` every second. Determine the magnitude and direction of the net acceleration of the cyclist on the circular turn.

Answer 1

Linear speed of the cyclist, `v=18xx(5)/(18)=5ms^(-1)`
and Centripetal acceleration the cyslist,
`a_(c)=(v^(2))/(R)=(25)/(25sqrt2)=(1)/(sqrt2)ms^(-2)`

Tangential acceleration of the cyclist, `a_(t)=(dv)/(dt)=-(1)/(2)ms^(-2)`
`therefore` Net acceleration of two cyclist,
`a_("net")sqrt(((1)/(sqrt2))^(2)+((1)/(2))^(2))=(sqrt3)/(2)=0.86ms^(-2)`
`tantheta=(a_(c))/(a_(t))=(1//sqrt2)/(1//2)=(2)/(sqrt2)=sqrt2`
Angle made by resultant acceleration with tangential acceleration,
`rArr" "theta=tan^(-1)(sqrt2)`