91Ó°ÊÓ

Displacement refers to the change in position of an object, measured as the shortest path between the initial and final points, typically along a straight line. For the pebble accelerating on a horizontal plane, the displacement provided in the problem is parallel to the x-axis and is given as 12.0 meters. It is important to distinguish between displacement and distance. While distance is the total path covered, displacement focuses only on the direct change in position.
In mathematical terms, displacement can be represented by the vector notation \( \Delta x \) for the x-axis and \( \Delta y \) for the y-axis, which shows the change in position from \( (x_0, y_0) \) to \( (x_1, y_1) \). For this problem, we are only concerned with the displacement along the x-axis.
Understanding displacement helps us to use kinematic equations effectively, which are used to predict future motion given initial conditions, such as velocity and acceleration. In problems involving displacement, these equations might help determine unknown quantities such as time or final velocity.

Constant acceleration means that an object's acceleration does not change over time. For the pebble in this scenario, it is subjected to a constant acceleration vector \( \vec{a} = (5.00 \, \text{m/s}^2) \hat{i} + (7.00 \, \text{m/s}^2) \hat{j} \). This constant acceleration implies that any change in velocity is consistent over equal time intervals.
Constant acceleration is a key concept in kinematics as it simplifies the equations of motion. These kinematic equations, such as \( v = v_0 + at \) and \( s = ut + \frac{1}{2}at^2 \), allow us to calculate unknowns like time, displacement, and velocity when certain parameters are known. More specifically, they help us solve problems involving linear motion in various directions.
In practice, to solve the problem, you would substitute the acceleration values into these equations to find out how the pebble's velocity changes over time. Using the given acceleration vector, the velocity components can be computed after identifying the time at which the displacement along the x-axis is 12 meters.

In two-dimensional motion, such as this, velocity is broken down into components along the x and y axes, also called velocity components. The initial velocity of the pebble is given as \( \vec{v}_0 = (4.00 \, \text{m/s}) \hat{i} + (0 \, \text{m/s}) \hat{j} \), with \( v_{0x} = 4.00 \, \text{m/s} \) and \( v_{0y} = 0 \, \text{m/s} \).
Kinematic equations aid in finding the final velocity components after a known time interval using constant acceleration. With the equation \( v = v_0 + at \), you can easily find the velocity in each direction. For the x-axis, substitute \( a_x = 5.00 \, \text{m/s}^2 \) and \( v_{0x} = 4.00 \, \text{m/s} \) to find \( v_x \). Similarly, use \( a_y = 7.00 \, \text{m/s}^2 \) and \( v_{0y} = 0 \, \text{m/s} \) for \( v_y \).
A complete solution will provide velocity components \( v_x \) and \( v_y \), typically combined to find the overall velocity magnitude, \( |\vec{v}| \), providing insight into the total movement.

Solving quadratic equations is a key mathematical tool needed in kinematics to find unknown variables like time \( t \). Quadratics arise when dealing with equations like \( s = ut + \frac{1}{2}at^2 \), often simplifying to a polynomial of the form \( at^2 + bt + c = 0 \).
In this problem, after substituting known values into the displacement equation \( x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2 \), a quadratic equation of \( 2.5t^2 + 4.00t - 12.0 = 0 \) emerges. To find \( t \), we use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where the coefficients \( a, b, \) and \( c \) are extracted from the equation.
By solving for \( t \), we find one or more possible solutions. Typically, the positive value is relevant in real-world scenarios where time cannot be negative. This value allows further calculation of final velocity components by applying them back into appropriate kinematic equations.