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  • 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.

Q. 4.31 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)

best_answer

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\ =\ \frac{v^2}{r}

                                                   =\ \frac{(7.5)^2}{80}

                                                   =\ 0.7\ m/s^2

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

So, the net acceleration becomes :

                                            a\ =\ \sqrt{a_c^2\ +\ a_T ^2}

or                                               =\ \sqrt{(0.7)^2\ +\ (0.5) ^2}

or                                              =\ 0.86\ m/s^2

Now for direction,   

                                         \tan \Theta \ =\ \frac{a_c}{a_T}

or                                                     =\ \frac{0.7}{0.5}

Thus,                                         \Theta \ =\ 54.46^{\circ} 

Posted by

Devendra Khairwa

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