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Chapter 2: Problem 37

Your friend's car starts from rest and travels \(0.500 \mathrm{~km}\) in \(10.0 \mathrm{~s}\). What is the magnitude of the constant acceleration required to do this?

Short Answer

Expert verified
Answer: The magnitude of the constant acceleration required for the car to travel 0.500 km in 10.0 seconds is \(10~\frac{m}{s^2}\).

Step by step solution

01

Write the given data

The car starts from rest, so its initial velocity \(v_0 = 0\). The distance traveled, \(d = 0.500~km\) which needs to be converted to meters: \(d = 500~m\). The time taken, \(t = 10.0~s\).
02

Use the equation for distance traveled under constant acceleration

We will use the equation for the distance traveled under constant acceleration: \(d = v_0t + \frac{1}{2}at^2\) Since the initial velocity is 0, we can simplify the equation to: \(d = \frac{1}{2}at^2\) Now, plug in the given values: \(500~m = \frac{1}{2}a(10.0~s)^2\)
03

Solve for acceleration, a

First, calculate the value inside the parentheses: \((10.0~s)^2 = 100~s^2\) Now, multiply both sides by 2: \(1000~m = a(100~s^2)\) Finally, divide both sides by 100 to find the acceleration: \(a = \frac{1000~m}{100~s^2} = 10~\frac{m}{s^2}\) The magnitude of the constant acceleration required for the car to travel 0.500 km in 10.0 s is \(10~\frac{m}{s^2}\).

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without taking into account the forces that cause the motion. It's about describing how objects move—think speed, velocity, and acceleration—in space and time.

Understanding kinematics is crucial for solving many physics problems, particularly those involving constant acceleration. As in the exercise, we're often interested in determining the displacement (change in position), velocity, and acceleration of objects over time. In the given problem, we were tasked with calculating the car's acceleration, using its displacement over a set period.
Equations of Motion
Equations of motion are mathematical formulas that describe the kinematic behavior of moving objects. There are a few key equations that are commonly used when an object is accelerating at a constant rate:
  • The first equation connects velocity, acceleration, and time: \(v = v_0 + at\).
  • The second equation associates displacement, initial velocity, and time: \(d = v_0t + \frac{1}{2}at^2\).
  • The third equation links displacement, velocity, and acceleration: \(v^2 = v_0^2 + 2ad\).
These equations are often referred to as the 'SUVAT' equations (from S = displacement, U = initial velocity, V = final velocity, A = acceleration, and T = time). By choosing the right equation based on the given and required variables, solving kinematics problems becomes more straightforward.
Acceleration Calculation
Acceleration is the rate at which an object's velocity changes over time. If this rate is constant, we say the object has constant acceleration. The formula to calculate constant acceleration when you know the displacement and time, as seen in our exercise example, is derived from the second SUVAT equation:\(d = \frac{1}{2}at^2\).

From this formula, we can solve for acceleration by rearranging the terms: \(a = \frac{2d}{t^2}\). It's crucial to plug in the correct units—usually meters for displacement and seconds for time—to get the acceleration in meters per second squared (\(\frac{m}{s^2}\)).
Physics Problem Solving
Physics problem solving is a systematic process that involves understanding the problem, identifying the relevant physical principles, choosing the appropriate mathematical tools, and carrying out the necessary calculations. Here's a simplified approach to solving physics problems:
  1. Read the problem carefully: Understand what is given and what needs to be found. In our example, we identified that the car's initial velocity, distance traveled, and time were given, and we needed to find the acceleration.
  2. Sketch the situation: Sometimes a simple diagram can help visualize the problem.
  3. Select the right formula: Based on the known and unknown variables, choose the equation that fits best.
  4. Plug in the known values: Substitute the numbers into the formula, with attention to units.
  5. Solve for the unknown: Rearrange the equation if necessary and perform the calculations to find the answer.
Following these steps can lead to successful problem-solving in physics, especially in kinematics and other areas.

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Most popular questions from this chapter

In 2005, Hurricane Rita hit several states in the southern United States. In the panic to escape her wrath, thousands of people tried to flee Houston, Texas by car. One car full of college students traveling to Tyler, Texas, 199 miles north of Houston, moved at an average speed of \(3.0 \mathrm{~m} / \mathrm{s}\) for one-fourth of the time, then at \(4.5 \mathrm{~m} / \mathrm{s}\) for another one-fourth of the time, and at \(6.0 \mathrm{~m} / \mathrm{s}\) for the remainder of the trip. a) How long did it take the students to reach their destination? b) Sketch a graph of position versus time for the trip.

In a fancy hotel, the back of the elevator is made of glass so that you can enjoy a lovely view on your ride. The elevator travels at an average speed of \(1.75 \mathrm{~m} / \mathrm{s}\). A boy on the 15th floor, \(80.0 \mathrm{~m}\) above the ground level, drops a rock at the same instant the elevator starts its ascent from the 1st to the 5th floor. Assume the elevator travels at its average speed for the entire trip and neglect the dimensions of the elevator. a) How long after it was dropped do you see the rock? b) How long does it take for the rock to reach ground level?

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). What maximum height above the projection point does it reach?

The planet Mercury has a mass that is \(5 \%\) of that of Earth, and its gravitational acceleration is \(g_{\text {mercury }}=3.7 \mathrm{~m} / \mathrm{s}^{2}\) a) How long does it take for a rock that is dropped from a height of \(1.75 \mathrm{~m}\) to hit the ground on Mercury? b) How does this time compare to the time it takes the same rock to reach the ground on Earth, if dropped from the same height? c) From what height would you have to drop the rock on Earth so that the fall- time on both planets is the same?

The position of a race car on a straight track is given as \(x=a t^{3}+b t^{2}+c,\) where \(a=2.0 \mathrm{~m} / \mathrm{s}^{3}, b=2.0 \mathrm{~m} / \mathrm{s}^{2}\), and \(c=3.0 \mathrm{~m}\). a) What is the car's position between \(t=4.0 \mathrm{~s}\) and \(t=9.0 \mathrm{~s}\) ? b) What is the average speed between \(t=4.0 \mathrm{~s}\) and \(t=9.0 \mathrm{~s} ?\)
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