A cyclist is riding with a speed of 27 km/h.
Question.
A cyclist is riding with a speed of 27 km/h. 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 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

solution:

$0.86 \mathrm{~m} / \mathrm{s}^{2} ; 54.46^{\circ}$ with the direction of velocity

Speed of the cyclist, $v=27 \mathrm{~km} / \mathrm{h}=7.5 \mathrm{~m} / \mathrm{s}$

Radius of the circular turn, $r=80 \mathrm{~m}$

Centripetal acceleration is given as:

$a_{\mathrm{c}}=\frac{v^{2}}{r}$

$=\frac{(7.5)^{2}}{80}=0.7 \mathrm{~m} / \mathrm{s}^{2}$

The situation is shown in the given figure:

Suppose the cyclist begins cycling from point $P$ and moves toward point $Q$. At point $Q$, he applies the breaks and decelerates the speed of the bicycle by $0.5$ $\mathrm{m} / \mathrm{s}^{2} .$

This acceleration is along the tangent at $Q$ and opposite to the direction of motion of the cyclist.

Since the angle between $a_{\mathrm{c}}$ and $a_{\mathrm{T}}$ is $90^{\circ}$, the resultant acceleration $a$ is given by:

Since the angle between $a_{\mathrm{c}}$ and $a_{\mathrm{T}}$ is $90^{\circ}$, the resultant acceleration $a$ is given by:

$a=\sqrt{a_{\mathrm{c}}^{2}+a_{\mathrm{T}}^{2}}$

$=\sqrt{(0.7)^{2}+(0.5)^{2}}$

$=\sqrt{0.74}=0.86 \mathrm{~m} / \mathrm{s}^{2}$

$\tan \theta=\frac{a_{\mathrm{c}}}{a_{\mathrm{T}}}$

Where $\theta$ is the angle of the resultant with the direction of velocity

$\tan \theta=\frac{0.7}{0.5}=1.4$

$\theta=\tan ^{-1}(1.4)$

$=54.46^{\circ}$
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