Until a train is a safe distance from the station it must travel at $$ 5 \mathrm{~m} / \mathrm{s} $$. Once the train is on open track it can speed up to $$ 45 \mathrm{~m} / \mathrm{s} $$. If it takes a train 8 seconds to reach $$ 45 \mathrm{~m} / \mathrm{s} $$, what is the acceleration of the train? (Round your answer to the nearest whole number.) $$ 4 \mathrm{~m} / \mathrm{s}^{2} $$ $$ 5 \mathrm{~m} / \mathrm{s}^{2} $$ $$ 6 \mathrm{~m} / \mathrm{s}^{2} $$ $$ 7 \mathrm{~m} / \mathrm{s}^{2} $$

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**Step 1:**  
Determine the change in speed. The train accelerates from an initial speed of $$ 5 \, \mathrm{m/s} $$ to a final speed of $$ 45 \, \mathrm{m/s} $$. Thus, the change in speed is  
$$
\Delta v = 45 \, \mathrm{m/s} - 5 \, \mathrm{m/s} = 40 \, \mathrm{m/s}.
$$

**Step 2:**  
Use the formula for acceleration, which is  
$$
a = \frac{\Delta v}{\Delta t}.
$$
Given the time interval is $$ 8 $$ seconds, substitute the values:  
$$
a = \frac{40 \, \mathrm{m/s}}{8 \, \mathrm{s}} = 5 \, \mathrm{m/s^2}.
$$

**Final Answer:**  
The acceleration of the train is $$ 5 \, \mathrm{m/s^2} $$.
The train's acceleration is $$ 5 \, \mathrm{m/s}^{2} $$.

Until a train is a safe distance from the station it must travel at . Once the train is on open track it can speed up to
. If it takes a train 8 seconds to reach , what is the acceleration of the train? (Round your answer to the
nearest whole number.)

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Until a train is a safe distance from the station it must travel at . Once the train is on open track it can speed up to . If it takes a train 8 seconds to reach , what is the acceleration of the train? (Round your answer to the nearest whole number.)