A cyclist is riding with a speed of 27 km/h. 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 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
Net acceleration is due to braking and centripetal acceleration
Due to Braking,
aT=0.5m/s2
Speed of the cyclist, v=27km/h=7.5m/s
Radius of the circular turn, r=80m
Centripetal acceleration is given as:
ac=V2r
=(7.52)/80=0.70m/s2
Since the angle between ac and aT is 900, the resultant acceleration a is given by:
a=(a2c+a2T)1/2
a=(0.72+0.52)1/2=0.86m/s2
tanθ=acaT
where θ is the angle of the resultant with the direction of the velocity.
tanθ=0.70.5=1.4
θ=tan−1(1.4)=54.560