91Ó°ÊÓ

The concept of constant speed is straightforward — it means moving at the same pace over time without speeding up or slowing down. In our problem, the motorist travels at a consistent velocity of \( 15.0 \mathrm{m/s} \). This uniform motion is described mathematically as \( v = \frac{d}{t} \), where \( v \) is the velocity, \( d \) is the displacement, and \( t \) is time.

To understand the significance of constant speed in kinematics, imagine you're watching a movie scene where a car cruises down a straight highway without changing its pace. This constant motion is what we see in our motorist's journey. The key takeaway is that when speed does not change, the distance covered (\(d\)) is directly proportional to time (\(t\)), resulting in the simple displacement equation \(d = v \cdot t \).

On the other side of constant speed is acceleration, which refers to the rate at which velocity changes. It can be an increase or decrease in speed, or a change in direction. The police officer in our scenario accelerates at \( 2.00 \mathrm{m/s^{2}} \), which is a constant acceleration. The formula for acceleration is typically written as \( a = \frac{\Delta v}{\Delta t} \), where \( a \) is acceleration, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the change in time.

In our problem, the police officer starts from rest, which means his initial speed (\( u \) ) is 0. His acceleration is constant, so we can use the equation \( v = u + a \cdot t \) to determine his velocity at any point in time. This acceleration will play a crucial role when calculating the time of overtaking, ensuring that he catches up with the motorist.

Displacement equations are the heart of solving kinematics problems as they relate velocity, acceleration, and time to the distance covered. In our textbook problem, there are two crucial displacement equations. The first is for the motorist's constant speed \( d_m = v_m \cdot t \), whereas the second is for the police officer who is accelerating, expressed as \( d_o = 0.5 \cdot a_o \cdot t^2 \).

It's key to note that displacement isn't the same as distance. Displacement is vector, which means it takes into account the direction. However, in this problem, since both the officer and the motorist are moving in the same direction along a straight line, we can work with the values as if they were distances. By equating \( d_m \) with \( d_o \) when the officer overtakes, we are able to solve for the time, as the problem dictates.

The time of overtaking is the moment when the accelerating body—in our case, the police officer—catches up with the body moving at constant speed—the motorist. To reach this critical point in our exercise, we need to find the instant when both the motorist and officer have covered the same displacement. Setting their displacement equations equal to one another and solving for time gives us the exact moment of overtaking.

One key improvement advice for understanding this concept is to visualize the situation. You can picture two dots on a line; one dot (motorist) moves at a steady pace, and the other (officer) starts from a standstill but begins to move faster and faster. Eventually, the accelerating dot catches up to the constant-speed dot; this intersection represents the time of overtaking. Using the derived equations, we found that it takes 7.5 seconds for the officer to reach the motorist.