Question

A car moving at \(60.0 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(t=4.00 \mathrm{~s}\). Assume uniform deceleration. a) How far does the car travel while stopping? b) What is its deceleration?

Short answer

(b) What is its deceleration? Answer: (a) The car travels 33.34 meters while stopping. (b) The deceleration of the car is -4.167 m/s².

Step-by-step solution

Identify the known quantities

We are given the following information: - Initial velocity, \(v_{i} = 60.0 \mathrm{~km/h}\) - Final velocity, \(v_{f} = 0 \mathrm{~km/h}\) (since the car comes to a stop) - Time, \(t = 4.00 \mathrm{~s}\) First, we'll convert the velocities from km/h to m/s, so that all measurements are in SI units: \(v_{i} = \frac{60.0 \times 1000}{3600} = 16.67 \mathrm{~m/s}\)

Find the distance traveled (a)

Using the first equation of motion, $$s = v_{i}t + \frac{1}{2}at^2$$ where s is the distance traveled, a is the deceleration, and t is the time. We have \(v_{i}\) and \(t\), but we'll need to find the deceleration a first.

Find the deceleration (b)

Using the second equation of motion, $$v_{f} = v_{i} + at$$ We know \(v_{f} = 0 \mathrm{~m/s}\), so we can rearrange for a: $$a = \frac{v_{f} - v_{i}}{t}$$ Substituting the values: $$a = \frac{0 - 16.67}{4.00} = -4.167 \mathrm{~m/s^2}$$ The deceleration of the car is \(-4.167 \mathrm{~m/s^2}\).

Calculate the distance traveled (a)

Now that we have the deceleration, we can go back to step 2 and plug in the values into the first equation of motion: $$s = v_{i}t + \frac{1}{2}at^2$$ $$s = (16.67) (4.00) + \frac{1}{2} (-4.167)(4.00)^2$$ $$s = 66.68 - 33.336 = 33.34 \mathrm{~m}$$ The car travels \(33.34 \mathrm{~m}\) while stopping.

Key concepts

Equations of Motion

When investigating physics problems involving motion, it's essential to understand the equations of motion that describe the relationship between velocity, acceleration, and displacement over time. These equations are vital tools for solving problems in kinematics, which is the study of motion without considering the forces that cause it.

In the exercise provided, the car's motion is described using such equations, specifically the first equation of motion: \[s = v_{i}t + \frac{1}{2}at^2\]where \(s\) represents the distance traveled, \(v_{i}\) is the initial velocity, \(a\) is the acceleration (or deceleration in this context), and \(t\) is the time elapsed. This equation is derived from the basic principles of kinematics and allows us to calculate how far an object moves when accelerating uniformly over a period of time.

Understanding the relationship between these variables is critical for breaking down motion problems into manageable steps. Therefore, it's a great approach to identify known quantities first and then apply the right equation to find the unknowns.

Kinematic Equations

Kinematic equations provide a mathematical description of motion with constant acceleration (or deceleration). These include four key formulas that relate displacement, initial and final velocities, acceleration, and time.

One of the kinematic equations was utilized in the provided exercise: \[v_{f} = v_{i} + at\]which is used to find the final velocity (\(v_{f}\)) given the initial velocity (\(v_{i}\)), acceleration (\(a\)), and time (\(t\)). For the scenario of the car coming to a stop, the final velocity is zero, and we can rearrange the equation to solve for the deceleration.

Deceleration as Negative Acceleration

Deceleration is simply negative acceleration, and thus the equation requires a negative sign when acceleration is in the direction opposite to the motion (slowing down). Encountering the term 'deceleration' instead of negative acceleration can aid learners in visualizing the direction of velocity changes.

Using these equations demands attention to units since they must all be consistent, typically in the SI system (meters, seconds, meters per second squared). It's beneficial to convert any non-SI units before proceeding with calculations to prevent errors and enhance comprehension.

Deceleration Calculation

Deceleration, or negative acceleration, is the rate at which an object decreases its velocity. Calculating it accurately is crucial for predicting how quickly a vehicle will come to a stop or for any scenario involving slowing down. The car problem exemplifies how deceleration comes into play in real-world contexts.

To compute deceleration, we use the rearranged kinematic equation: \[a = \frac{v_{f} - v_{i}}{t}\]Substituting the car's initial velocity and the time it takes to stop into the equation allows one to find the uniform deceleration. It's important to remember that deceleration values are expressed with a negative sign to signify the reduction in speed.

Deceleration and Distance

Once the deceleration is known, it can be used to determine other aspects of motion, such as the distance the car travels while stopping. This is done by inserting the deceleration value back into the first equation of motion. The careful balance between conceptual understanding and mathematical application is key to mastering physics problems.

In conclusion, incorporating these concepts thoroughly aids in grasping the full picture of how objects move and allows for a more intuitive approach to problem-solving in physics.