To find the magnitude of the train's acceleration, we can use the formula:
\[ a = \frac{\Delta v}{\Delta t} \]
Where:
- \( a \) is the acceleration,
- \( \Delta v \) is the change in velocity,
- \( \Delta t \) is the change in time.
- Convert the final speed from kilometers per hour to meters per second.
- The final speed is 144 kilometers per hour.
- To convert to meters per second, use the conversion factor \( \frac{1000 , \text{meters}}{1 , \text{kilometer}} \) and \( \frac{1 , \text{hour}}{3600 , \text{seconds}} \):
\[ 144 , \text{km/h} = 144 \times \frac{1000}{3600} \approx 40 , \text{m/s} \]
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Calculate the change in velocity.
- The train started from rest, so \( \Delta v = 40 , \text{m/s} - 0 , \text{m/s} = 40 , \text{m/s} \).
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The time taken for the acceleration was 180 seconds.
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Substituting into the acceleration formula:
\[ a = \frac{40 , \text{m/s}}{180 , \text{s}} \approx 0.222 , \text{m/s}^2 \approx 0.22 , \text{m/s}^2 \]
Therefore, the information the engineer should convey to the crew is:
The magnitude of acceleration was 0.22 m/s².