A train is traveling at $55 \text{miles per hour}$ and begins to brake to a stop. If the average deceleration is constant at $0.5 m/s^2$, approximately how long does it take for the train to come to a complete stop?

• A. $49 s$
• B. $49 \min$
• C. $49 s$
• D. $110 s$

Answer: (C)

Verified step by step guidance

1. Convert the initial speed of the train from miles per hour (mph) to meters per second (m/s) because the deceleration is given in meters per second squared. Use the conversion factor: 1 mile = 1609.34 meters and 1 hour = 3600 seconds. The formula is: \(v_0 = 55 \times \frac{1609.34}{3600}\) m/s.
2. Identify the known quantities: initial velocity \(v_0\) (converted to m/s), final velocity \(v = 0\) m/s (since the train comes to a stop), and constant deceleration \(a = -0.5\) m/s\(^2\) (negative because it is slowing down).
3. Use the kinematic equation that relates velocity, acceleration, and time: \(v = v_0 + a t\). Since the final velocity \(v\) is zero, rearrange the equation to solve for time \(t\): \(t = \frac{v - v_0}{a}\).
4. Substitute the known values into the equation for \(t\) using the converted initial velocity and the given acceleration to find the time it takes for the train to stop.
5. Interpret the result as the duration in seconds for the train to come to a complete stop under the given deceleration.