91Ó°ÊÓ

Kinematic equations are fundamental to understanding motion in physics, especially in scenarios involving constant acceleration. These equations allow us to predict and calculate various aspects of a moving object, including its position, velocity, and acceleration at any given time. For a car accelerating from rest, the most relevant kinematic equation is:

• \[ s = u t + \frac{1}{2} a t^2 \]

where \(s\) represents the distance traveled, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. In this exercise, the second car uses this equation to determine how far it travels over a specific time period. Because it starts from rest, the initial velocity \(u = 0\) eliminates the \(u t\) term, simplifying the equation to \[ s = \frac{1}{2} a t^2 \]. By rearranging this equation, we can find the required acceleration to meet another moving car at a specific point.

Relative motion describes the movement of two or more objects in relation to each other. In this exercise, two cars are moving on the highway with one trying to catch up with the other. The first car maintains a constant speed of \(33 \mathrm{m/s}\), while the second car accelerates to meet it at the next exit. The key to solving such problems is understanding that both cars must travel the same distance in the same amount of time to meet at the specified location. This involves comparing their movements using kinematic equations to ensure their paths cross under the given conditions. Relative motion becomes especially important when coordinating the dynamics of different moving objects, like when the second car accelerates to coincide with the steady speed of the first car.

Calculating time and distance accurately is crucial in physics problems involving motion, such as the scenario with the two cars. The first step is often to convert all units to a consistent system, such as converting 2.5 km to 2500 meters. For constant speed, the time \(t\) to travel a known distance \(d\) can be determined using the formula:

• \[ t = \frac{d}{v} \]

where \(v\) is the velocity. In our case, the first car's time to the exit is calculated to show how long both vehicles have to meet at the exit. With this time, we then apply it to the kinematic equation of the second car to solve for the acceleration. This highlights the interconnectedness of time and distance in determining motion parameters, ensuring both cars meet at the exit simultaneously. The calculations take advantage of constant velocity for the first car and constant acceleration for the second, incorporating both aspects of time and distance effectively.