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When an object is accelerating at a constant rate, it is said to be in uniformly accelerated motion. This concept is incredibly useful in physics, as many dynamic systems in both natural and engineered scenarios exhibit this kind of motion. Let's consider a subway train that starts from rest. This means that initially, its speed is zero. However, the train doesn’t just jump to its cruising speed instantaneously; it accelerates over time.
In the case of our subway train, it accelerates at a rate of 1.60 m/s² for 14.0 seconds. The acceleration is uniform or constant, hence the term 'uniformly accelerated motion'. Understanding this concept allows us to predict how an object will behave over time when subjected to a constant force. Additionally, it's the basis for several important physics equations that enable us to calculate variables such as distance, velocity, and time during such motion.

The equations of motion are a set of formulas that describe the relationship between velocity, acceleration, distance, and time. They are essential for solving problems in kinetics, like our subway train scenario. There are four primary equations, but in the solution provided, we use two of them:

• To calculate velocity: \( v = u + at \), where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
• For the distance: \( s = ut + \frac{1}{2}at^2 \) where \( s \) is distance, and the other variables are as previously defined.

These equations assume uniformly accelerated motion and no other forces acting on the object besides the acceleration (like air resistance). By applying these equations, we can analyse each phase of the train’s journey and calculate distances covered during acceleration, constant velocity, and deceleration.

Calculating the distance an object travels over a period of time is a core part of physics. In our example, we examine a train moving through three distinct phases: acceleration, constant speed, and deceleration. Each phase requires a different approach:

• During acceleration, the distance is found using \( s = ut + \frac{1}{2}at^2 \), where initial speed \( u \) is zero.
• At a constant speed, the formula simplifies to \( s = vt \), where \( v \) is the constant speed and \( t \) is the time.
• The deceleration phase has a similar formula to acceleration, but deceleration is used instead.

By calculating distances for each phase separately and then combining them, we get the total distance traversed by the train during its journey from one station to the next.

Deceleration, often also referred to as negative acceleration, is the rate at which an object slows down. It is the same as acceleration in terms of calculation but with a crucial difference: acceleration increases the velocity of an object, while deceleration decreases it. In the train's case, after running at a constant speed for a period, it needs to slow down to stop at the next station.
To find out how long it takes to stop (time of deceleration), we can use the equation \( t = \frac{v - u}{a} \), considering that final velocity \( v \) will be zero when the train stops, initial velocity \( u \) is the velocity at the end of the constant speed phase, and \( a \) is the deceleration rate. Once we have the time, we use the distance formula \( s = ut + \frac{1}{2}at^2 \) to determine the distance covered during deceleration. This insight is critical for safety in real-world situations such as vehicular traffic, where the ability to slow down efficiently can be lifesaving.