Amtrak's 20 th-Century limited is en route from Chicago to New York at \(110 \mathrm{km} / \mathrm{h}\) when the engineer spots a cow on the track. The train brakes to a halt in 1.2 min, stopping just in front of the cow. (a) What is the magnitude of the train's acceleration? (b) What's the direction of the acceleration? (c) How far was the train from the cow when the engineer applied the brakes?

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It involves studying the various aspects of motion such as position, velocity, and acceleration. When we look at the train acceleration problem, kinematics allows us to describe the train's motion from the moment the brakes are applied to when it comes to a halt.

Kinematics problems usually involve calculations based on formulas that define the relationships between these properties of motion. Common kinematic equations relate initial velocity, final velocity, acceleration, time, and distance covered. In the case of the train in our exercise, knowing the initial speed, the fact that it comes to a stop, and the time taken to stop, we can apply kinematic equations to find out both the acceleration of the train and the distance covered during braking.

Uniformly accelerated motion refers to motion in which an object experiences a constant acceleration. This means that the velocity of the object changes at a steady rate. For the 20th-Century limited train solving uniformly accelerated motion problems involves a set of kinematic equations applied under the assumption that the acceleration is consistent throughout the motion. In our textbook example, the train decelerates at a uniform rate—which is another way of saying it has a constant negative acceleration. This constant rate allows us to predict the duration it will take to come to a stop and how far it will travel in that time.

The equation we use for calculating the uniform acceleration is given by the change in velocity over the change in time, denoted as \(a = \frac{{\Delta v}}{{\Delta t}}\). Here, applying this equation to the train's situation, where it reduces from initial velocity to zero, helps us find the acceleration with which the train slows down. The uniform nature of the acceleration simplifies the calculations and gives us a clear picture of the train's deceleration.

Deceleration is simply negative acceleration; it's when an object slows down. The magnitude of deceleration tells us how quickly the velocity of the object is decreasing. In kinematic terms, it's still acceleration, just with a negative sign to indicate the object is losing speed rather than gaining it. In our train acceleration problem, deceleration is what happens as the train brakes in response to the engineer spotting a cow on the track.

The direction of deceleration is always opposite to the direction of motion. For the train, which initially moves forward, deceleration occurs in the reverse direction, effectively reducing the train's speed until it comes to a stop. It's crucial to factor in this negative sign when calculating the distance covered during deceleration, as it influences the calculation of the overall displacement. So, for our train, that is stopping, the deceleration can be calculated by the change in velocity over time, producing a negative acceleration value indicating slowing down.