The distance, \(d,\) an accelerating object travels in \(t\) seconds can be modeled by the equation \(d=\frac{1}{2} a t^{2},\) where \(a\) is the acceleration rate, in meters per second per second. If a car accelerates from a stop at the rate of 20 meters per second per second and travels a distance of 80 meters, about how many seconds did the car travel? A. Between 1 and 2 B. Between 2 and 3 C. Between 3 and 4 D. 4 E. 8

Understanding object acceleration is crucial when studying motion in physics. Acceleration is defined as the rate of change of velocity per unit of time. In simpler terms, it tells us how quickly an object is speeding up or slowing down. The units of acceleration are typically meters per second squared (m/s²).

In our exercise, a car is accelerating from a complete stop, which means its initial velocity is 0 m/s. The acceleration, denoted as 'a', is given as 20 m/s². This implies that every second, the speed of the car increases by 20 meters per second. When analyzing motion, it's important to recognize that constant acceleration results in the distance traveled increasing quadratically over time—this is where our kinematic equation comes into play, representing the distance-time relationship under constant acceleration. The equation is written as:
\[d = \frac{1}{2} a t^2\].

This equation is derived from integrating the acceleration with respect to time to find velocity, and then integrating velocity to find distance. It's crucial to understand that the factor of \(\frac{1}{2}\) comes from these integrations and reflects the increasing distance covered as the object speeds up.

The distance-time relationship in kinematics is a foundational concept that describes how an object's position changes over time. In the context of constant acceleration, this relationship is not linear, meaning that distance doesn't increase in a straight line as time goes on. Instead, the distance is proportional to the square of the elapsed time, which is expressed in the kinematic equation:
\[d = \frac{1}{2} a t^2\].

In our exercise, the distance-time relationship is applied to a car accelerating from rest. Given that the car travels 80 meters, we want to know the time it took to travel this distance under constant acceleration. Using our equation, the relationship between the distance and the time is quadratic, highlighting that the car covers more distance in each successive time interval. This quadratic relationship is what allows us to solve for the time variable and understand how distance increases with the square of time when acceleration is constant. The comprehension of this concept is vital, as it helps predict how the distance covered by an accelerating object evolves with time.

Physics problem solving often involves a systematic approach to dissecting the question and applying relevant equations and concepts. In dealing with kinematics problems, like the one we're examining, the following steps are typically employed:

• Understand the problem and visualize the scenario.
• Identify the known and unknown quantities, along with the appropriate units.
• Choose the right equation(s) connecting these quantities.
• Rearrange the equation(s) if necessary to isolate the unknown variable.
• Substitute the known values into the equation and solve for the unknown.
• Reflect on the answer to ensure it makes sense in the context of the question.

In this problem, we followed these steps to isolate 't' in the distance-time relationship equation and substituted the known values to find the time it took for the car to travel 80 meters. Problem-solving skills in physics are honed by practice and applying these steps consistently. Additionally, understanding the underlying concepts rather than just memorizing formulas is essential for solving problems efficiently and accurately.