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A bank robber in a getaway car approaches an intersection at a speed of 45 mph. Just as he passes the intersection, he realizes that he needed to turn. So he steps on the brakes, comes to a complete stop, and then accelerates driving straight backward. He reaches a speed of \(22.5 \mathrm{mph}\) moving backward. Altogether his deceleration and reacceleration in the opposite direction take \(12.4 \mathrm{~s} .\) What is the average acceleration during this time?

Short Answer

Expert verified
Answer: The average acceleration during this time is approximately -2.433 m/s².

Step by step solution

01

Convert speeds to m/s

First, to work with the speeds in SI units, we need to convert the given speeds from miles per hour (mph) to meters per second (m/s). We'll use the conversion factor 1 mile = 1609.34 m, and 1 hour = 3600 seconds. Initial speed: \[45~\mathrm{mph} * \frac{1609.34~\mathrm{m}}{1~\mathrm{mile}} * \frac{1~\mathrm{hour}}{3600~\mathrm{s}} = 20.1168~\mathrm{m/s}\] Final backward speed: \[22.5~\mathrm{mph} * \frac{1609.34~\mathrm{m}}{1~\mathrm{mile}} * \frac{1~\mathrm{hour}}{3600~\mathrm{s}} = 10.0584~\mathrm{m/s}\] Note that the robber moves backward, making the final speed negative: \[-10.0584~\mathrm{m/s}\]
02

Calculate the acceleration

Now that we have both the initial and final velocities in m/s, we can calculate the average acceleration using the formula: \[a_{avg} = \frac{vf - vi}{t}\] Where: \(a_{avg}\) is the average acceleration, \(vf\) is the final velocity, \(vi\) is the initial velocity, \(t\) is the time.
03

Plug in the values and solve

We'll now plug in the values we've found into the formula to find the average acceleration: \(a_{avg} = \frac{-10.0584~\mathrm{m/s} - 20.1168~\mathrm{m/s}}{12.4~\mathrm{s}}\) \(a_{avg} = \frac{-30.1752~\mathrm{m/s}}{12.4~\mathrm{s}}\) \(a_{avg} = -2.433~\mathrm{m/s^2}\) So, the average acceleration during this time is approximately \(-2.433~\mathrm{m/s^2}\).

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Key Concepts

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

Physics Problem Solving
Understanding how to tackle physics problems is essential for students looking to master the subject. In the context of this exercise, the key elements of problem-solving include interpreting the scenario, identifying knowns and unknowns, choosing the right formula, unit conversion, and plugging in the values.

To begin, students must carefully read the problem to comprehend the physical situation. Here, a bank robber's change in velocity is crucial to finding the average acceleration. Identifying that there is a deceleration followed by an acceleration in the opposite direction is also essential. Selecting the correct physics formula comes next, which in this case is the average acceleration formula: \(a_{avg} = \frac{vf - vi}{t}\).

The practice of solving problems in physics also hinges on the correct handling of units which requires converting mph to m/s in this problem. By making sure the units are consistent—using the SI system in this case—students avoid calculation errors that can distort the solution. Always double-check each step to ensure precision and understanding.
Kinematics
Kinematics is the branch of physics that deals with motion without considering the forces that cause the motion. It involves analyzing the movement of objects using concepts like velocity, acceleration, displacement, and time. In the problem at hand, the kinematics principle applied is the calculation of average acceleration, which provides a way to understand how the velocity of an object changes over time.

Average acceleration is defined as the change in velocity divided by the time during which the change occurred. This can be visualized on a velocity-time graph where acceleration corresponds to the slope of the line connecting the initial and final velocities. Kinematic equations streamline the process of solving physics problems by offering a straightforward means to calculate unknown quantities when certain variables are given. Students should familiarize themselves with these formulas and the conditions under which they are applicable as they are fundamental to understanding motion.
Velocity Conversion
Velocity conversion is a practical skill in physics, particularly when comparing different systems of measurement. The problem we're discussing demonstrates a typical conversion from miles per hour to meters per second, which is essential for consistent calculations within the SI unit system.

To perform such conversions, one needs to know the relevant conversion factors. Here, the following are used: 1 mile = 1609.34 meters, and 1 hour = 3600 seconds. When converting, you multiply the given velocity by these factors to transform mph into m/s. For instance:
  • Initial velocity: \(45 \mathrm{mph} \times \frac{1609.34 \mathrm{m}}{1 \mathrm{mile}} \times \frac{1 \mathrm{hour}}{3600 \mathrm{s}} = 20.1168 \mathrm{m/s}\)
  • Final backward velocity: \(22.5 \mathrm{mph} \times \frac{1609.34 \mathrm{m}}{1 \mathrm{mile}} \times \frac{1 \mathrm{hour}}{3600 \mathrm{s}} = 10.0584 \mathrm{m/s}\) (made negative due to the direction of travel)
Memorizing these factors isn't essential, but understanding the conversion process is. Practice with different values to grasp the concept better and ensure accuracy.
SI Units
SI units, or the International System of Units, is the most widely used system for measurements in science. This standardized system ensures consistency across various fields and helps in the clear communication and understanding of scientific and technical data. In mechanics, the fundamental SI units include meters (m) for length, kilograms (kg) for mass, and seconds (s) for time.

In the context of the acceleration problem, speed initially provided in mph (miles per hour) is not an SI unit, which is why conversion to m/s (meters per second) is necessary. Likewise, acceleration is expressed in m/s², translating to meters per second squared. Always use SI units when solving physics problems unless otherwise directed, and convert any given measurements into SI units before applying equations. This practice avoids errors and facilitates simpler, more accurate calculations.

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Most popular questions from this chapter

A car traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) applies the brakes and decelerates uniformly at a rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) a) How far does it travel in \(3.0 \mathrm{~s}\) ? b) What is its velocity at the end of this time interval? c) How long does it take for the car to come to a stop? d) What distance does the car travel before coming to a stop?

An F-14 Tomcat fighter jet is taking off from the deck of the USS Nimitz aircraft carrier with the assistance of a steam-powered catapult. The jet's location along the flight deck is measured at intervals of \(0.20 \mathrm{~s} .\) These measurements are tabulated as follows: $$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|} \hline t(\mathrm{~s}) & 0.00 & 0.20 & 0.40 & 0.60 & 0.80 & 1.00 & 1.20 & 1.40 & 1.60 & 1.80 & 2.00 \\ \hline x(\mathrm{~m}) & 0.0 & 0.70 & 3.0 & 6.6 & 11.8 & 18.5 & 26.6 & 36.2 & 47.3 & 59.9 & 73.9 \\ \hline \end{array} $$ Use difference formulas to calculate the jet's average velocity and average acceleration for each time interval. After completing this analysis, can you say if the F- 14 Tomcat accelerated with approximately constant acceleration?

You and a friend are standing at the edge of a snowcovered cliff. At the same time, you throw a snowball straight upward with a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) over the edge of the cliff and your friend throws a snowball straight downward over the edge of the cliff with the same speed. Your snowball is twice as heavy as your friend's. Neglecting air resistance, which snowball will hit the ground first, and which will have the greater speed?

A ball is dropped from the roof of a building. It hits the ground and it is caught at its original height 5.0 s later. a) What was the speed of the ball just before it hits the ground? b) How tall was the building? c) You are watching from a window \(2.5 \mathrm{~m}\) above the ground. The window opening is \(1.2 \mathrm{~m}\) from the top to the bottom. At what time after the ball was dropped did you first see the ball in the window?

An object is thrown vertically upward and has a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

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