Question

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 !

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Answer 1

Net acceleration is due to braking and centripetal acceleration

Due to Braking,

a =0.5m/s²

T

Speed of the cyclist, v=27km/h=7.5m/s

Radius of the circular turn, r=80m

Centripetal acceleration is given as:

a = V²/r

c

=(7.5²)/80=0.70m/s²

Since the angle between a and a is 90°,

c T

the resultant acceleration a is given by:

a=(a² + a² ) ½

c T

a=(0.7² +0.5²) ½ = 0.86m/s²

tanθ= a / a

c T

where θ is the angle of the resultant with the direction of the velocity.

tanθ= 0.7/0.5= 1.4

θ=tan -¹(1.4) = 54.56°