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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of${\mathbf{216}}{\mathbf{km}}{\mathbf{/}}{\mathbf{h}}$ . (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to ${\mathbf{0}}{\mathbf{.}}{\mathbf{050}}{\mathbf{g}}$, what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a ${\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathbf{km}}$ radius to be at the acceleration limit?

a) Smallest radius of curvature of the trackis $7.3km$.

b) Maximum allowed speed of the train is$80km/h$.

See the step by step solution

## Step 1: Given data

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

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

## Step 2: Understanding the concept of centripetal acceleration

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.

## Step 3: Conversion of units

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

So,

$\mathrm{v}=216\frac{\mathrm{km}}{\mathrm{h}}\phantom{\rule{0ex}{0ex}}=216\frac{\mathrm{km}}{\mathrm{h}}×\frac{1000\mathrm{m}}{3600\mathrm{s}}\phantom{\rule{0ex}{0ex}}=60\mathrm{m}/\mathrm{s}$

## Step 4: (a) Calculating the smallest radius of curvature

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

$=\frac{{\left(60\mathrm{m}/\mathrm{s}\right)}^{2}}{0.050×9.8\mathrm{m}/{\mathrm{s}}^{2}}\phantom{\rule{0ex}{0ex}}=7347\mathrm{m}\phantom{\rule{0ex}{0ex}}=7.3\mathrm{km}$

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

## Step 5: (b) Calculating the required speed for 1.00 km radius of curvature

Now, the radius of curvature available is given as,

$\mathrm{r}=1\mathrm{km}\phantom{\rule{0ex}{0ex}}=1000\mathrm{m}$

${a}_{c}=\frac{{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}{v}^{2}={a}_{c}×r$

Substitute the values,

$=0.050×9.8\mathrm{m}/{\mathrm{s}}^{2}×1000\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{v}=22.1\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\approx 80\mathrm{km}/\mathrm{h}$

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

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