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An object moving with uniform acceleration has a velocity of \(12.0 \mathrm{cm} / \mathrm{s}\) in the positive \(x\) direction when its \(x\) coordinate is \(3.00 \mathrm{cm} .\) If its \(x\) coordinate \(2.00 \mathrm{s}\) later is \(-5.00 \mathrm{cm},\) what is its acceleration?

Short Answer

Expert verified
The acceleration of the object is -8.5 cm/s^2.

Step by step solution

01

Identify Known Variables

First, identify what information you have available. The initial position (x) is 3.00 cm, the final position (x') is -5.00 cm, the initial velocity (v) is 12.0 cm/s, and the time interval (t) is 2.00 s. We are looking for the acceleration (a).
02

Use the appropriate equation

Now we use the equation from kinematics that involves all the quantities we know, and the one we are trying to find. The equation is \(x' = x + vt+0.5at^2\).
03

Substitute known values into equation

Substitute all known values into the equation, which yields \(-5.00 cm = 3.00 cm + (12.0 cm/s * 2.00 s)+0.5a(2.00 s)^2\).
04

Solve for acceleration (a)

Now you need to solve the equation for acceleration. This yields \(a = \frac{-5.00 cm - 3.00 cm - (12.0 cm/s * 2.00 s)}{0.5*(2.00 s)^2}\). After calculating, you get \(a = -8.5 cm/s^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
Kinematic equations are fundamental in understanding the motion of objects when that motion is uniform in acceleration. These equations describe how the position of an object changes with time. They are derived from basic principles of acceleration and velocity, and they play a crucial role in solving problems in physics.

There are four main kinematic equations - each linking time, displacement, initial velocity, acceleration, and final velocity in different ways. The key point is that they assume acceleration is constant, which means the object does not speed up or slow down at varying rates during the period of time considered. For the exercise at hand, we selected the equation:
\[ x' = x + vt + \frac{1}{2}at^2 \]
where:
  • \( x' \) is the final position,
  • \( x \) is the initial position,
  • \( v \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( t \) is the time.
The appropriate equation is chosen based on the information given and what needs to be found. In the example problem, all but the acceleration are known, making it possible to solve for acceleration by rearranging the equation.
Initial Velocity
Initial velocity refers to the speed of an object at the start of a time interval. In kinematic problems, it's important to know the direction and magnitude of the initial velocity because it affects how the object moves over time. Initial velocity can be positive or negative, depending on the direction of motion.

The given exercise states that the initial velocity is \(12.0\,\text{cm/s}\) in the positive \(x\) direction. This information is crucial for setting up the kinematic equation correctly. If the initial velocity is not set properly, the entire calculation can go wrong. Understanding the effects of initial velocity on motion can help students visualize how an object will travel from one point to another over a period, considering consistent acceleration.
Acceleration Calculation
Acceleration is the rate at which an object changes its velocity. It is a vector quantity, meaning it has both magnitude and direction. To calculate acceleration in a kinematic context, we usually rearrange a kinematic equation to solve for \(a\). As indicated in the exercise, after substituting known values from the problem into the equation, acceleration can be isolated and solved for.

The calculation will look like this:
\[ a = \frac{x' - x - vt}{0.5t^2} \]
By substituting the known values for \(x', x, v,\) and \(t\), you obtain:
\[ a = \frac{-5.00\,\text{cm} - 3.00\,\text{cm} - (12.0\,\text{cm/s} \times 2.00\,\text{s})}{0.5\times (2.00\,\text{s})^2} \]
Through this calculation, we find that the acceleration is \(-8.5\,\text{cm/s}^2\), indicating the object is slowing down in the positive \(x\) direction. Displaying this process clearly helps students understand how acceleration affects an object's motion over time and can improve their problem-solving skills.

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