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 ?
Given: the speed of the cyclist is 27 km / h , the radius of the turn is 80 m and deceleration of the cyclist is 0.50 m / s 2 .
Let θ be the angle of the net acceleration with the tangential velocity, v be the tangential velocity of the cyclist, a c be the centripetal acceleration, a T be the tangential acceleration of the cyclist.
The figure shows the motion of the cyclist at point P.
The centripetal acceleration is,
a c = v 2 R
Substitute the values in the above expression.
a c = ( 27 × 1000 60 × 60 ) 2 80 = 45 64 m / s
The centripetal acceleration a c and the tangential acceleration a t act perpendicular to each other.
The net acceleration is given by,
a = a c 2 + a t 2
Substitute the values in the above expression.
a = ( 45 64 ) 2 + ( 0.50 ) 2 = 0.86 m / s 2
From the figure, the tangent of the angle is,
tan θ = a c a t
Substitute the values in the above expression.
tan θ = 45 64 × 0.50 θ = tan − 1 ( 45 32 ) = 54.5 °
Thus, the magnitude and direction of the net acceleration of the cyclist are 0.86 m / s 2 and 54.5 ° .
Given: the speed of the cyclist is
Let
The figure shows the motion of the cyclist at point P.
The centripetal acceleration is,
Substitute the values in the above expression.
The centripetal acceleration
The net acceleration is given by,
Substitute the values in the above expression.
From the figure, the tangent of the angle is,
Substitute the values in the above expression.
Thus, the magnitude and direction of the net acceleration of the cyclist are
