$\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} $

$\qquad= [(0.4)^2+(0.3) ^2]^{1/2}$

$\qquad= 0.5 \;ms^{-2}$

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