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(I) A sports car accelerates from rest to \(95 \mathrm{~km} / \mathrm{h}\) in \(4.5 \mathrm{~s}\). What is its average acceleration in \(\mathrm{m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The average acceleration is approximately 5.86 m/s².

Step by step solution

01

Convert Final Velocity to Meters per Second

The sports car reaches a speed of 95 km/h. To calculate acceleration in meters per second squared (m/s²), convert this speed into meters per second (m/s). Use the conversion factor: 1 km/h = 0.27778 m/s. Therefore, 95 km/h is equivalent to \( 95 \times 0.27778 \approx 26.39 \) m/s.
02

Determine the Change in Velocity

The car starts from rest, meaning its initial velocity (\( v_0 \)) is 0 m/s. The change in velocity (\( \Delta v \)) is the final velocity minus the initial velocity. Thus, \( \Delta v = 26.39 \text{ m/s} - 0 \text{ m/s} = 26.39 \text{ m/s} \).
03

Calculate the Average Acceleration

Acceleration is defined as the change in velocity divided by the time taken to change that velocity. Use the formula: \( a = \frac{\Delta v}{t} \), where \( \Delta v = 26.39 \) m/s and \( t = 4.5 \) s. Therefore, the average acceleration is \( a = \frac{26.39}{4.5} \approx 5.86 \text{ m/s}^2 \).

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Key Concepts

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

Conversion of Units
In physics, especially dealing with motion, it's crucial to ensure that units are consistent across calculations. This is known as the conversion of units. In the exercise above, we started with a velocity given in kilometers per hour (km/h). However, to determine acceleration in meters per second squared (m/s²), it's important to convert this velocity into meters per second (m/s).

The conversion factor between these units is 1 km/h equals approximately 0.27778 m/s. Here's how you can convert:
  • Take the initial velocity in km/h.
  • Multiply it by 0.27778 to find its equivalent in m/s.
For instance, 95 km/h converts to about 26.39 m/s, making it simpler to compute acceleration using this common system of units in Kinematics.
Kinematics Equations
Kinematics is the branch of physics that deals with the motion of objects without considering their masses or the forces that cause the motion. When an object changes its velocity over time, as the sports car did, it experiences acceleration.

To determine the average acceleration, we employ kinematics equations. One of the primary equations used is the formula for acceleration:\[ a = \frac{\Delta v}{t}\]This equation states that acceleration (\(a\)) is the change in velocity (\(\Delta v\)) divided by the time period (\(t\)) over which this change occurs.
  • Kinematic equations are useful for calculating unknown variables such as distance, final velocity, or time when other variables are known.
  • They rely on knowing initial conditions and consistent units to provide accurate results.
Understanding how to manipulate and apply these equations allows one to analyze and predict the motion of objects efficiently in various circumstances.
Velocity Change
Velocity is a vector quantity that includes both speed and direction. In the exercise, the sports car's velocity changes from 0 to a final value. Calculating this change is crucial for finding acceleration.

The **change in velocity**, denoted as \(\Delta v\), is simply the final velocity minus the initial velocity. In the case given:
  • Initial velocity (\(v_0\)) = 0 m/s (since the car starts from rest).
  • Final velocity (\(v_f\)) = 26.39 m/s (after conversion from km/h).
Thus, the change in velocity \(\Delta v\) is 26.39 m/s. This value then feeds into the kinematic equation for acceleration, allowing for a straightforward calculation of how quickly the car gains speed over the given time frame.

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Most popular questions from this chapter

(II) A horse canters away from its trainer in a straight line, moving \(116 \mathrm{~m}\) away in \(14.0 \mathrm{~s}\). It then turns abruptly and gallops halfway back in 4.8 s. Calculate \((a)\) its average speed and \((b)\) its average velocity for the entire trip, using "away from the trainer" as the positive direction.

(I) What must your car's average speed be in order to travel \(235 \mathrm{~km}\) in \(3.25 \mathrm{~h} ?\)

(II) A stone is thrown vertically upward with a speed of \(24.0 \mathrm{~m} / \mathrm{s}\). \((a)\) How fast is it moving when it reaches a height of \(13.0 \mathrm{~m} ?(b)\) How much time is required to reach this height? (c) Why are there two answers to (b)?

(III) Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration: $$ a=\frac{d v}{d t}=g-k v $$ where \(k\) is a constant. ( \(a\) ) Derive a formula for the velocity of the body as a function of time assuming it starts from rest \((v=0\) at \(t=0)\). [Hint: Change variables by setting \(u=g-k v .](b)\) Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.

A fugitive tries to hop on a freight train traveling at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\). Just as an empty box car passes him, the fugitive starts from rest and accelerates at \(a=1.2 \mathrm{~m} / \mathrm{s}^{2}\) to his maximum speed of \(6.0 \mathrm{~m} / \mathrm{s} .(a)\) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car?

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