Question

# A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution a distance of $25\,{\text{m}}$ along the circumference of the circle in $5.0\,{\text{s}}$. The magnitude of her acceleration is:A. $0.31\,{\text{m/}}{{\text{s}}^2}$B. $1.3\,{\text{m/}}{{\text{s}}^2}$C. $1.6\,{\text{m/}}{{\text{s}}^2}$D. $3.9\,{\text{m/}}{{\text{s}}^2}$

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Hint:Use the formula for centripetal acceleration of an object. This formula gives the relation between the velocity of the object and radius of the circular path. Hence, calculate the velocity of the girl using her displacement and time required for the displacement. Also calculate the radius of the circle equating the displacement of the girl with one fourth of the circumference of the circle.

Formulae used:
The centripetal acceleration $a$ of an object in the circular motion is given by
$a = \dfrac{{{v^2}}}{R}$ …… (1)
Here, $v$ is the velocity of the object and $R$ is the radius of the circular path.
The velocity $v$ of an object is
$v = \dfrac{s}{t}$ …… (2)
Here, $s$ is the displacement of the object and $t$ is the time.

We have given that a girl is jogging on the circumference of the horizontal circle. She covers the distance of $25\,{\text{m}}$ which is one fourth of the circumference of the horizontal circle in time $5.0\,{\text{s}}$.
$s = 25\,{\text{m}}$
$\Rightarrow t = 5.0\,{\text{s}}$
Let us first calculate the velocity of the girl.Substitute $25\,{\text{m}}$ for $s$ and $5.0\,{\text{s}}$ for $t$ in equation (2).
$v = \dfrac{{25\,{\text{m}}}}{{5.0\,{\text{s}}}}$
$\Rightarrow v = 5\,{\text{m/s}}$
Hence, the velocity of the girl is $5\,{\text{m/s}}$.

Let us now calculate the radius of the circular track.The girl jogs only on one fourth of the total circumference of the horizontal circle. Hence, the displacement of the girl is equal to the one fourth of the circumference of the horizontal circle.
$s = \dfrac{1}{4}\left( {2\pi R} \right)$
$\Rightarrow R = \dfrac{{2s}}{\pi }$
Substitute $25\,{\text{m}}$ for $s$ and $3.14$ for $\pi$ in the above equation.
$\Rightarrow R = \dfrac{{2\left( {25\,{\text{m}}} \right)}}{{3.14}}$
$\Rightarrow R = 15.9\,{\text{m}}$
Hence, the radius of the circular track is $15.9\,{\text{m}}$.

We can calculate the acceleration of the girl using equation (1).Substitute for $5\,{\text{m/s}}$ $v$ and $15.9\,{\text{m}}$ for $R$ in equation (1).
$a = \dfrac{{{{\left( {5\,{\text{m/s}}} \right)}^2}}}{{15.9\,{\text{m}}}}$
$\Rightarrow a = 1.57\,{\text{m/}}{{\text{s}}^2}$
$\therefore a \approx 1.6\,{\text{m/}}{{\text{s}}^2}$
Therefore, the magnitude of acceleration of the girl is $1.6\,{\text{m/}}{{\text{s}}^2}$.

Hence, the correct option is C.

Note:The students may take the formula for the acceleration of the object as ratio of velocity to time. But here we have asked to calculate the magnitude of centripetal acceleration of the girl and the formula for centripetal acceleration is not the ratio of the velocity and time. Hence, the students should be careful while using the formula for the acceleration in this case.