91Ó°ÊÓ

1) Average speed of the train is$216km/h$

2) Acceleration experience by the passengers is limited to$0.050g$

When an object moves in a circular orbit, it will experience a centripetal acceleration. The magnitude of this acceleration depends on the speed of the object as well as the radius of curvature. Using the formula for centripetal acceleration, we can find the radius of the curvature of the track if the velocity is known, or we can find the velocity if the radius of curvature is known.

Formula:

$a_c=\frac{V^2}{r}$

Where $a_c$is the centripetal acceleration, $v$is the average speed, and $r$is the radius of curvature.

First, we need to convert the units of given speed from km/h to m/s

So,

$$\begin{gathered} \mathrm{v}=216\frac{\mathrm{km}}{\mathrm{h}} \\ =216\frac{\mathrm{km}}{\mathrm{h}}×\frac{1000\mathrm{m}}{3600\mathrm{s}} \\ =60\mathrm{m}/\mathrm{s} \end{gathered}$$

To determine the smallest radius of curvature, we need to use formula forcentripetal acceleration.

$a_c=\frac{V^2}{r}$

Rearranging,

$r=\frac{v^2}{a_c}$

Substitute $60\mathrm{m}/\mathrm{s}$the forv, and$9.8\mathrm{m}/{\mathrm{s}}^2$for g in the above equation.

$$\begin{gathered} =\frac{{\left(60\mathrm{m}/\mathrm{s}\right)}^2}{0.050×9.8\mathrm{m}/{\mathrm{s}}^2} \\ =7347\mathrm{m} \\ =7.3\mathrm{km} \end{gathered}$$

Therefore, the smallest radius of curvature is$7.3\mathrm{km}$ .

Now, the radius of curvature available is given as,

$$\begin{gathered} \mathrm{r}=1\mathrm{km} \\ =1000\mathrm{m} \end{gathered}$$

$$\begin{gathered} a_c=\frac{v^2}{r} \\ {v}^2=a_c×r \end{gathered}$$

Substitute the values,

$$\begin{gathered} =0.050×9.8\mathrm{m}/{\mathrm{s}}^2×1000\mathrm{m} \\ \mathrm{v}=22.1\mathrm{m}/\mathrm{s} \\ ≈80\mathrm{km}/\mathrm{h} \end{gathered}$$

Hence, the speed at the curvature should be approximately $80\mathrm{km}/\mathrm{h}$.