91Ó°ÊÓ

When analyzing the motion of objects, we rely on the equations of motion. These equations describe the mathematical relationships between displacement, velocity, acceleration, and time. For objects moving with constant acceleration, as in our exercise scenario, we have a few specific equations that help us understand their movement.
The first equation concerns the displacement of an object starting from rest. If an object begins moving with constant acceleration, its position over time can be expressed as:

• \( x(t) = \frac{1}{2} a t^2 \)

This equation shows us that position changes quadratically with time when starting from rest.
In our problem, this equation defines car 1's position as it starts accelerating towards car 2. For car 2, moving with constant velocity, its position is governed by a linear equation:

• \( x(t) = D - v_0 t \)

This tells us that car 2's position decreases linearly with time because it is moving toward car 1 at a steady speed.

Constant acceleration occurs when the change in velocity over time remains the same. This is a crucial concept in kinematics, as it simplifies the calculation of an object's future position and velocity.
Constant acceleration means that every second, an object's velocity increases by the same amount. In car 1's case, if it starts from rest, its acceleration \( a_x \) adds progressively more to its speed. The velocity of car 1 over time can be expressed by the equation:

• \( v(t) = a_x t \)

This linear relationship indicates that as time increases, so does the velocity, resulting in the characteristic parabolic path described by its position function.
Meanwhile, car 2 moves with a constant velocity \( v_0 \), meaning there is no acceleration acting on it. This stark contrast between the motion of the two cars makes understanding constant acceleration critical for analyzing the entire problem.

The quadratic equation is a crucial tool in solving kinematics problems where variable relationships involve squared terms. In this exercise, the collision time is where the motion paths of the two cars converge. We find this by setting the position equations equal, leading to a quadratic equation:

• \( \frac{1}{2} a_x t^2 + v_0 t - D = 0 \)

To solve this, we use the quadratic formula:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, coefficients are defined as \( a = \frac{1}{2} a_x \), \( b = v_0 \), and \( c = -D \).
By applying this formula, we determine the time \( t \) when both cars meet. Only the positive root is considered since time cannot be negative in our scenario. Hence, this quadratic equation provides us the exact collision time, given the cars' initial states and car 1's acceleration.

The velocity-time graph provides a visual depiction of how an object's speed changes over time. For car 1, which starts from rest and accelerates, its velocity \( v_x(t) = a_x t \) results in a straight line with a positive slope in a velocity-time graph.
As time progresses, this slope shows that the velocity increases linearly with time. By finding the point on this graph corresponding to the collision time, we can easily read off car 1’s velocity at the moment of impact.
In contrast, car 2's velocity-time graph is a horizontal line at \( -v_0 \), indicating its constant velocity. There is no slope since its speed doesn't change over time. This pair of graphs helps us understand and visualize the differing dynamics of each car's movement throughout the exercise.