A cylclist is riding with a speed of `27 km h^-1`. 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 `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

A cylclist is riding with a speed of `27 km h^-1`. As he approaches a

1.	A cylclist is riding with a speed of `27 km h^-1`. 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 `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
Answer» Correct Answer - `tan^-1(5//7)` with radius. `v = 27 xx (5)/(18) = (15)/(2) ms^-1, a_c = ((15//2)2)/(80) = 0.7 ms^-2` `a_t = 0.5 ms^-2` Net acceleration : `a = sqrt(a_c^2 + a_t^2) = 0.86 ms^-2` Direction `tan theta = (a_t)/(a_c) = (0.5)/(0.7)` `theta = tan^-1(5//7)` with radius.

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A cylclist is riding with a speed of `27 km h^-1`. 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 `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

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