Question

During an Olympic bobsled run, the Jamaican team makes a turn of radius ${}^{\mathbf{7}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{m}}$at a speed of ${}^{\mathbf{96}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$. What is their acceleration in terms of${}^{\mathbf{g}}$?

Short answer

The acceleration in terms of g is${}^{9.64g}$.

Step-by-step solution

Given

$$\begin{gathered} v-96.6\mathrm{km}/\mathrm{h} \\ -26.8\mathrm{m}/\mathrm{s} \\ \mathrm{R}=7.6\mathrm{m} \end{gathered}$$

Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity.

Formula:

The acceleration is given by,

$a=\frac{v^2}{R}$

where, v is the velocity, a is an acceleration and R is the radius.

Determining the acceleration in terms of g

$$\begin{gathered} a=\frac{v^2}{R} \\ =\frac{{\left(26.8\mathrm{m}/\mathrm{s}\right)}^2}{7.6\mathrm{m}} \\ =94.5\mathrm{m}/\mathrm{s} \end{gathered}$$

which can be expressed as a multiple of g,

$$\begin{gathered} a-\left(\frac{a}{g}\right)g \\ =\left(\frac{94.7\mathrm{m}/{\mathrm{s}}^2}{}\right)g \\ =9.64g \end{gathered}$$

Hence, the acceleration in terms of g is 9.64g.