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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. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Answers (1)

D Deependra Verma

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 .

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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