a cyclist riding with a speed of 27kmph.As he approaches a circular turn on the road of radius 80m, he applies breaks and reduces his speed at the constant rate of 0.50m/s every second. the net acceleration of cyclist on the circular turn is
Answer (1)
28th Sep, 2020
Dear student,
Due to Braking,
At = 0.5 m/s^2 ( At = Tangential acceleration )
Speed of the cyclist,v=27 km/hr = 275/18 = 7.5 m/s
Radius of the circular turn, r=80m
Centripetal acceleration (Ac) is given as:
Ac = v^2/r
Ac = ((7.5)^2)/80 = 0.7 m/s^2
Since the angle between Ac and At is 90degrees
Therefore, the resultant or net acceleration (A) is given by :-
A = (Ac^2 + At^2)^0.5
A = (0.7^2+0.5^2)^ 0.5 = 0.86 m/s^2
Due to Braking,
At = 0.5 m/s^2 ( At = Tangential acceleration )
Speed of the cyclist,v=27 km/hr = 275/18 = 7.5 m/s
Radius of the circular turn, r=80m
Centripetal acceleration (Ac) is given as:
Ac = v^2/r
Ac = ((7.5)^2)/80 = 0.7 m/s^2
Since the angle between Ac and At is 90degrees
Therefore, the resultant or net acceleration (A) is given by :-
A = (Ac^2 + At^2)^0.5
A = (0.7^2+0.5^2)^ 0.5 = 0.86 m/s^2
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