Question

A train slows down as it rounds a sharp horizontal turn, going from$\mathbf{90}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{km/h}}$to$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{km/h}}$ in the$\mathbf{15}\mathbf{}\mathbf{s}$it takes to round the bend. The radius of the curve is$\mathbf{\text{150m}}$. Compute the acceleration at the moment the train speed reaches$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{km/h}}$. Assume the train continues to slow down at this time at the same rate.

Short answer

The acceleration of the train at the moment its speed reaches$50.0\text{km}/\text{h}$ is$1.48 {\text{m/s}}^{\text{2}}$ inward and backward $\text{29.9}°$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of the train decreases from 90.0 km/h to 50.0 km/h .
• The radius of the curve is 150 m.

The tangential acceleration

The formula to calculate the tangential acceleration is,

${\mathit{a}}_{\mathbf{\text{t}}}\mathbf{=}\mathbf{-}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}$ ...(1)

Here,

${\mathit{a}}_{\mathbf{\text{t}}}$is the tangential acceleration.

$\mathbf{dv}$is the change in the velocity.

The tangential acceleration

The change in the velocity is calculated as,

$dv=v_1-v_2$

Here,

$v_1$is the initial velocity of the train during sharp horizontal turn.

$v_2$is the final velocity of the train after crossing the horizontal turn.

Substitute 90km/h for v₁ and 45km/h for v₂ in the equation

$\begin{array}{rcl}dv& =& 90.0 \text{km/h}-50.0 \text{km/h}\\ & =& 40.0 \text{km/h}\\ & & \end{array}$

Substitut the values in equation 1 and we get,

$\begin{array}{rcl}{a}_{\text{t}}& =& -\frac{40.0 \text{km/h}}{15 \text{s}}\\ & =& -\frac{40.0 \text{km/h}\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)\left(\frac{1 \text{h}}{3600 \text{s}}\right)}{15 \text{s}}\\ & =& -0.74 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Thus, the tangential acceleration is$\begin{array}{rcl}& & -0.74 {\text{m/s}}^{\text{2}}\end{array}$ .

The tangential acceleration is negative it means the acceleration is inward.

The radial acceleration of the train

The formula to calculate the radial acceleration of the train is,

$a_{\text{r}}=\frac{v^2}{r}$

Here,

r is the radius of the path.

v is the velocity of the train at the instant.

a_r is the radial acceleration.

Substitute values in the above equation.

$\begin{array}{rcl}{a}_{\text{r}}& =& \frac{{(50.0 \text{km/h})}^2}{150\text{m}}\\ & =& \frac{{(50.0 \text{km/h})}^2{\left(\frac{{10}^3 \text{m}}{1 \text{km}}\right)}^2{\left(\frac{1 \text{h}}{3600 \text{s}}\right)}^2}{150 \text{m}}\\ & =& 1.286 {\text{m/s}}^{\text{2}}\\ & \approx & 1.29 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Thus, the radial acceleration of the train is $\begin{array}{rcl}& & 1.29 {\text{m/s}}^{\text{2}}\end{array}$.

The relation between the total acceleration, tangential acceleration and the radial acceleration

The relation between the total acceleration, tangential acceleration and the radial acceleration is,

$a^2={\left(a_{\text{t}}\right)}^2+{\left(a_{\text{r}}\right)}^2$

Here,

a is the total acceleration.

Substitute the values in the above expression and we get,

$$\begin{gathered} a^2={\left(-0.74 {\text{m/s}}^2\right)}^2+{\left(1.29 {\text{m/s}}^{\text{2}}\right)}^2 \\ a=\sqrt{2.2117} {\text{m/s}}^{\text{2}} \\ a=1.4871 {\text{m/s}}^{\text{2}} \\ a\approx 1.48 {\text{m/s}}^{\text{2}} \end{gathered}$$

Thus, the total acceleration is$1.48 {\text{m/s}}^{\text{2}}$ inward.

The direction of the acceleration

The direction of the acceleration is calculated as,

$$\begin{gathered} \tan \theta =\frac{\left|a_{\text{t}}\right|}{\left|a_{\text{r}}\right|} \\ \theta ={\tan }^{-1}\left(\frac{\left|a_{\text{t}}\right|}{\left|a_{\text{r}}\right|}\right) \end{gathered}$$

Here,

$\theta$is the direction of the acceleration.

Substitute the values in the above expression, and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{\left|-0.74 {\text{m/s}}^{\text{2}}\right|}{\left|1.29 {\text{m/s}}^2\right|}\right) \\ ={\tan }^{-1}\left(\frac{0.74}{1.29}\right) \\ =29.840° \\ \approx 29.9° \end{gathered}$$

Thus, the direction of the acceleration is$29.9°$ backward.

Therefore, the acceleration of the train at the moment its speed reaches 50 km/h is $1.48 {\text{m/s}}^{\text{2}}$inward and $29.9°$backward.