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For a standard production car, the highest road-tested acceleration ever reported occurred in \(1993,\) when a Ford RS200 Evolution went from zero to \(26.8 \mathrm{~m} / \mathrm{s}(60 \mathrm{mi} / \mathrm{h})\) in \(3.275 \mathrm{~s}\). Find the magnitude of the car's acceleration.

Short Answer

Expert verified
The car's acceleration is approximately 8.18 m/s².

Step by step solution

01

Identify Initial and Final Speeds

The initial speed of the car is 0 m/s because it starts from rest, and the final speed is given as 26.8 m/s.
02

Identify the Time Interval

The time taken for the car to reach the final speed of 26.8 m/s is given as 3.275 seconds.
03

Use the Definition of Acceleration

Acceleration is defined as the change in velocity divided by the time taken to change that velocity. The formula to find acceleration \(a\) is: \[ a = \frac{\Delta v}{\Delta t} \] where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time.
04

Calculate the Change in Velocity

Since the initial velocity \(v_i = 0\) m/s and the final velocity \(v_f = 26.8\) m/s, the change in velocity \(\Delta v\) is: \[ \Delta v = v_f - v_i = 26.8 - 0 = 26.8 \text{ m/s} \]
05

Plug Values into Acceleration Formula

Using the acceleration formula, plug in the values of \(\Delta v\) (change in velocity) and \(\Delta t\) (time). Thus, the acceleration \(a\) is: \[ a = \frac{26.8 \, \text{m/s}}{3.275 \, \text{s}} \approx 8.18 \, \text{m/s}^2 \]
06

Conclude the Calculation

After performing the division, the magnitude of the acceleration is found to be approximately 8.18 m/s².

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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 physics that describes the motion of objects without considering the causes of the motion. In essence, it focuses on the geometry of motion, such as displacement, velocity, and acceleration. When we talk about any moving object, like a car accelerating down a track, kinematics helps us understand how the object's speed and position change over time.

To describe motion in kinematics, we often need to consider:
  • Position: Where an object is located at a particular time. It's usually described using coordinates in space.
  • Velocity: How fast an object changes its position. It tells us both the speed and direction of motion.
  • Acceleration: The rate at which an object changes its velocity. It can involve speeding up, slowing down, or changing direction.
Kinematics is simplified greatly when we consider linear motion, where all movement is in a straight line. This makes calculations straightforward, like in the example of the Ford RS200 Evolution, where acceleration was calculated using simple kinematic equations.
Acceleration Calculation
Acceleration is a key concept in physics, especially when discussing motion. It tells us how an object speeds up, slows down, or changes direction. Calculating acceleration involves understanding the change in velocity over a period of time.

To calculate acceleration, we can use the formula:
  • \[ a = \frac{\Delta v}{\Delta t} \]
Here, \( a \) is acceleration, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the change in time.

In our example of the Ford RS200 Evolution:
  • The initial velocity \( v_i \) was 0 m/s (since the car started from rest).
  • The final velocity \( v_f \) was 26.8 m/s.
  • The time \( \Delta t \) taken to reach this velocity was 3.275 seconds.
This information allows us to use the acceleration formula straightforwardly, resulting in an acceleration of approximately 8.18 m/s². Remember, the magnitude of acceleration is always positive when describing how fast something speeds up or slows down.
Motion Equations
In the study of motion, we use motion equations to predict and analyze the behavior of moving objects. These equations link together key variables of motion: displacement, velocity, acceleration, and time.

For linear motion, such as a car moving in a straight line, some key equations include:
  • Equation for velocity: \( v = u + at \) where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
  • Equation for displacement: \( s = ut + \frac{1}{2}at^2 \) where \( s \) is displacement.
  • Equation linking velocity and displacement: \( v^2 = u^2 + 2as \).
These equations allow us to solve a wide range of motion-related problems, such as calculating how far a car travels before coming to a stop or how long it takes to reach a particular velocity.

In our example, we specifically used the basic equation for acceleration because we were interested in how quickly the Ford RS200 Evolution accelerated to a certain speed. Understanding these equations empowers us to unravel the complexities of motion and make predictions about how objects move.

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Most popular questions from this chapter

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed of \(1.8 \mathrm{~m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{~m}\) (measured from the board), assuming that the downward direction is chosen as the negative direction. \(]\) (b) What is the highest point he reaches above the water?

(a) In general, does the average acceleration of an object have the same direction as its initial velocity \(v_{0}\), its final velocity \(v,\) or the difference \(v-v_{0}\) between its final and initial velocities? Provide a reason for your answer. (b) The following table lists four pairs of initial and final velocities for a boat traveling along the \(x\) axis. Use the concept of acceleration presented in Section 2.3 to determine the direction (positive or negative) of the average acceleration for each pair of velocities. $$ \begin{array}{|c|c|c|} \hline & \text { Initial velocity } v_{0} & \text { Final velocity } v \\ \hline(\mathrm{a}) & +2.0 \mathrm{~m} / \mathrm{s} & +5.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{b}) & +5.0 \mathrm{~m} / \mathrm{s} & +2.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{c}) & -6.0 \mathrm{~m} / \mathrm{s} & -3.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{d}) & +4.0 \mathrm{~m} / \mathrm{s} & -4.0 \mathrm{~m} / \mathrm{s} \\ \hline \end{array} $$ Problem The elapsed time for each of the four pairs of velocities is \(2.0 \mathrm{~s}\). Find the average acceleration (magnitude and direction) for each of the four pairs. Be sure that your directions agree with those found in the Concept Questions.

A whale swims due east for a distance of \(6.9 \mathrm{~km},\) turns around and goes due west for \(1.8 \mathrm{~km},\) and finally turns around again and heads \(3.7 \mathrm{~km}\) due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale?

A log is floating on swiftly moving water. A stone is dropped from rest from a 75 -mhigh bridge and lands on the log as it passes under the bridge. If the log moves with a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\), what is the horizontal distance between the \(\log\) and the bridge when the stone is released?

A jogger accelerates from rest to \(3.0 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s}\). A car accelerates from 38.0 to \(41.0 \mathrm{~m} / \mathrm{s}\) also in \(2.0 \mathrm{~s}\). (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the \(2.0 \mathrm{~s}\) ? If so, how much farther?

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