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?

Answers (1)

Speed of cycle =   27 Km/h  =  7.5 m/s

The situation is shown in figure :-  

                             Motion in plane ,         20244

The centripetal acceleration is given by :

                                           a_c = fracv^2r

                                                   = frac(7.5)^280

                                                   = 0.7 m/s^2

And the tangential acceleration is given as  0.5 m/s^2.

So, the net acceleration becomes :

                                            a = sqrta_c^2 + a_T ^2

or                                               = sqrt(0.7)^2 + (0.5) ^2

or                                              = 0.86 m/s^2

Now for direction,   

                                         	an Theta  = fraca_ca_T

or                                                     = frac0.70.5

Thus,                                         Theta  = 54.46^circ 

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