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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A cyclist is moving with a speed of $6 ms^{-1} $, As he approaches a circular turn on the road of radius $120 \;m$ he applies brakes and reduces his speed at a constant rate of $ 0.4 \;ms^{-2} $. The magnitude of the net acceleration of the cyclist on the circular turn is

$\begin{array}{1 1} 0.5\; ms^{-2} \\ 1.0\;ms^{-2} \\ 2.0\;ms^{-2} \\ 4.0 \;ms^{-2} \end{array} $

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Answer :$ 0.5\; ms^{-2}$
From the fig, the cyclist is moving on a straight road from A to B with a velocity $v= 6\;ms^{-1} $
As he approaches the circular turn, he decelerates at rate $a_r$ , represented by vector BD.
The magnitude of declaration is $ a_1= 0.4 \;ms^{-2}$
At point B, two accelerations $a_r$ and $a_c$ the centripetal acceleration directed towards the center C act on the cyclist.
Now $a_c=\large\frac{v^2}{R}$
$\qquad= \large\frac{(6)^2}{120}$
$\qquad= 0.3 \;ms^{-2}$
Using the law of parallelogram of vector addition, vector BE gives the resultant acceleration a whose magnitude is $ (DE= a_c)$
$a= (a_r^2+a_c^2 )^{1/2} $
$\qquad= [(0.4)^2+(0.3) ^2]^{1/2}$
$\qquad= 0.5 \;ms^{-2}$
answered Aug 25, 2014 by meena.p

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