A Cyclist is riding with a speed of 27 km/h.As he approaches a circular turn on the radius 80m,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 accelaration of the cyclist on the circular turn?
Asked by aruni_maiyer | 26th Sep, 2010, 01:17: PM
In the question, there are two accelerations involved. First one is the acceleration due to the breaks whose value is 0.5m/s/s and second is the due to the circular motion of the turn and its value is v*v/r, where v is the speed and r is the radius of the turn. Now , the acceleration of the breaks is along the turn and the circular acceleration is towards the centre, so both these accelerations are perpendicular to each other and net acceleration will be given by the resultant of these.
Now, speed of the cycle = 27km/h = 27000/3600 = 7.5 m/s
so, the value of centripetal acceleration = 7.5*7.5/80 = 0.703125
So, net acceleration = sqrt(0.703125*0.703125+ 0.5*0.5) = 0.86277 m/s/s.
The direction of this accelertion will be towards inwards and at an angle of tan-inverse 1.4 from the circular path.
Answered by | 26th Sep, 2010, 01:38: PM
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