Problem 28 - Fundamentals of Physics

Question

On a dry road, a car with good tires may be able to brake with a constant decclcration of \(4.92 \mathrm{~m} / \mathrm{s}^{2}\). (a) How long docs such a car, initially traveling at \(24.6 \mathrm{~m} / \mathrm{s}\), take to stop? (b) How far does it travel in this time? (c) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the deceleration.

Step-by-Step Solution

Answer

(a) 5.0 seconds; (b) 61.5 meters; (c) Draw graphs: parabolic for position, linear for velocity.

Step 1: Understand the Problem

We need to calculate the time it takes for a car traveling initially at a velocity of 24.6 m/s to come to a complete stop under a constant acceleration (deceleration in this case) of -4.92 m/s². After finding the time, we calculate the distance it travels during this time. Finally, we will graph the position and velocity as functions of time.

Step 2: Calculate the Time to Stop

To find how long it takes for the car to stop, we use the equation for acceleration: \[ v_f = v_i + a t \] Here, the final velocity \(v_f = 0\) m/s, initial velocity \(v_i = 24.6\) m/s, and acceleration \(a = -4.92\) m/s².Rearranging the equation for time \(t\): \[ t = \frac{v_f - v_i}{a} = \frac{0 - 24.6}{-4.92} \approx 5.0 \text{ seconds} \] So, it takes approximately 5.0 seconds to stop.

Step 3: Calculate the Stopping Distance

To find the distance the car travels before stopping, we use the equation: \[ s = v_i t + \frac{1}{2} a t^2 \] Substituting the known values, \(v_i = 24.6\) m/s, \(a = -4.92\) m/s², and \(t = 5.0\) s:\[ s = 24.6 \times 5.0 + \frac{1}{2} \times (-4.92) \times (5.0)^2 \] \[ s = 123 - 61.5 = 61.5 \text{ meters} \] The car travels 61.5 meters before coming to a stop.

Step 4: Graph Position and Velocity vs. Time

To create the graphs: - **Position vs. Time**: Starts at 0 and increases reaching 61.5 m at 5 s. It is a parabolic curve opening downwards because of the constant negative acceleration. - **Velocity vs. Time**: Starts at 24.6 m/s and decreases linearly to 0 m/s at 5 s. The slope of the line is the constant deceleration of -4.92 m/s². These two graphs provide a visualization of how the car's motion changes over time.

Key Concepts

Constant Acceleration

Constant acceleration occurs when the rate of change of velocity is unchanging over time. It means the velocity increases or decreases by the same amount each second. This simplifies calculations in physics because we can use straightforward equations that account for this constant rate.

In our exercise, the car is experiencing constant acceleration, but in the form of deceleration. The car initially moves at 24.6 m/s and, due to the brakes, slows down with an acceleration of \(-4.92 \text{ m/s}^2\). The negative sign indicates that the acceleration is reducing the car's speed.

Key equations for scenarios with constant acceleration include:

• \(v_f = v_i + at\): Final velocity \(v_f\) depends on initial velocity \(v_i\), acceleration \(a\), and time \(t\).
• \(s = v_i t + \frac{1}{2} a t^2\): Distance \(s\) traveled, with initial velocity \(v_i\) and time \(t\), influenced by acceleration \(a\).

Constant acceleration is a cornerstone concept in kinematics, allowing for predictable calculations in motion problems.

Deceleration

Deceleration is simply acceleration with a negative sign, indicating a reduction in speed. It's what happens when an object slows down. Any decrease in velocity over time is termed deceleration.

In our problem, the car decelerates at \(-4.92 \text{ m/s}^2\). This implies that every second, the car's speed decreases by 4.92 m/s. As a result, the car reaches a stop over a calculated time of approximately 5 seconds.

Key aspects of deceleration include:

• Like acceleration, it is measured in \(\text{m/s}^2\).
• It is experienced in everyday scenarios, such as braking in a vehicle.

Deceleration is crucial to understand in motion-related problems, helping us determine how quickly an object can come to rest when influenced by a force that reduces speed.

Motion Graphs

Motion graphs are visual representations of how an object's motion changes over time. They can feature displacement, velocity, and acceleration plotted against time, revealing patterns that equations describe numerically.

In the context of our problem, we are interested in two types of motion graphs:

• **Position vs. Time**: This graph shows how far the car travels over time. It starts at zero, representing the beginning position, and peaks at 61.5 meters when the car stops. The curve is parabolic due to the constant deceleration, demonstrating how the car's position changes at an increasingly slower rate.
• **Velocity vs. Time**: This graph begins at 24.6 m/s, the initial speed, and decreases linearly to 0 m/s in 5 seconds. The slope is representative of our constant acceleration, or more accurately, the steady deceleration.

Motion graphs provide a clear and intuitive way to understand and predict an object's behavior in motion. They translate complex calculations into simple visual insights.