A train pulls out of a station accelerating at \(0.1 \mathrm{m} / \mathrm{s}^{2}\). What is its speed, in kilometers per hour, \(21 / 2\) minutes after leaving the station?

Acceleration is a key concept in understanding how objects change their velocity over time. It is defined as the rate of change of velocity per unit time. In the given problem, the train accelerates at a constant rate of 0.1 m/s². This means for every second, the train's velocity increases by 0.1 meters per second. Understanding this concept is crucial because it connects the time the train has been moving to its final speed.
For example, if a train continuously accelerates for 1 second at 0.1 m/s², it gains a velocity of 0.1 m/s. If it accelerates for 10 seconds, its speed will increase to 1 m/s. The formula to calculate the speed when acceleration and time are known is:
\[v = a \times t\]
where:

• v = final velocity (m/s)
• a = acceleration (m/s²)
• t = time (seconds)

Applying this understanding to the problem, we use the formula to find the speed after 630 seconds.

Unit conversion is vital for solving physics problems effectively, as it ensures consistency in measurements. In this exercise, we need to convert time from minutes to seconds and speed from meters per second to kilometers per hour. Let's discuss how to perform these conversions:

• **Time Conversion:** Given time is 21/2 minutes. To convert from minutes to seconds, we use the fact that 1 minute = 60 seconds. The conversion is: \[10.5 \times 60 = 630\] seconds.
• **Speed Conversion:** To convert the speed from m/s to km/h, we need to know that 1 kilometer = 1000 meters and 1 hour = 3600 seconds. Therefore, the conversion factor from m/s to km/h is \[\frac{3600}{1000} = 3.6\]. Thus, the speed in km/h is obtained by multiplying the speed in m/s by 3.6.

So, a speed of 63 m/s is converted to kilometers per hour by: \[63 \times 3.6 = 226.8\] km/h.

Kinematics equations are extremely useful tools in physics that relate the variables of motion: displacement, velocity, acceleration, and time. In this exercise, we use the kinematics equation for velocity under constant acceleration: \[v = a \times t\].
This equation assumes that the initial velocity (

• v₀) is zero, as the train starts from rest. If there were an initial velocity, the equation would be:
• v = v₀ + a \times t

In general, kinematics equations help us to solve a wide variety of problems involving motion. It's essential to understand which equation to apply in a given scenario.
The main kinematics equations include:

• s = v₀ × t + 0.5 × a × t² (equation for displacement)
• v = v₀ + a × t (equation for final velocity)
• v² = v₀² + 2a × s (equation relating velocity and displacement)

Each of these equations serves a different purpose, depending on the variables involved in the problem. Understanding these can help solve complex problems by breaking them down into manageable steps.