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Chapter 2: Problem 43

a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? b. What fraction of \(g\) is this? c. How far has the car traveled when it reaches 60 mph? Give your answer both in SI units and in feet.

Short Answer

Expert verified
The car must have a constant acceleration of approximately \(2.68224 m/s^2\), which is about \(0.274\) of \(g\). The car would have traveled about \(134.112 m\) or \(439.963 ft\) when it reaches \(60 mph\).

Step by step solution

01

Convert Speed from mph to m/s

Make sure all units are in SI units for ease of calculation. The given final speed of the car is 60 mph. Convert this to metres per second (m/s) using the conversion factor \(1 mph = 0.44704 m/s\). Therefore,\( 60 mph = 60 * 0.44704 m/s = 26.8224 m/s\).
02

Apply First Equation of Motion

The first equation of motion is \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is acceleration and \(t\) is time. Here, \(u = 0 m/s\) because the car starts from rest, \(v = 26.8224 m/s\) and \(t = 10s\). Solve the equation for \(a\): \(a = (v-u) / t = 26.8224 m/s 梅 10s = 2.68224 m/s^2\).
03

Express Acceleration as a Fraction of Gravitational Constant

The acceleration due to gravity \(g\) is approximately \(9.8 m/s^2\). Compute \(a/g\) to get the fraction of acceleration to gravity: \(a/g = 2.68224 m/s^2 梅 9.8 m/s^2 \approx 0.274\). Thus, the acceleration is approximately 0.274 of the gravitational constant.
04

Calculate the Displacement Using Second Equation of Motion

Using the second equation of motion \(s = ut + 0.5at^2\), where \(s\) is the displacement, compute for \(s\): \(s = ut + 0.5at^2 = 0 (as u = 0) + 0.5 x 2.68224 m/s^2 x (10 s)^2 = 134.112 m\).
05

Convert Displacement from Meters to Feet

Use the conversion factor \(1 m = 3.281 ft\) to convert \(134.112 m\) to feet: \(134.112 m * 3.281 ft/m = 439.963 ft\).

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Key Concepts

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

Units Conversion
When solving physics problems, one must often convert units to ensure consistency in equations. For velocity, a common conversion is from miles per hour (mph) to meters per second (m/s), because m/s is the standard in the International System of Units (SI). To do this, use the conversion factor:
  • 1 mph = 0.44704 m/s
For example, to convert 60 mph to m/s:
  • Calculate: 60 mph 脳 0.44704 m/s = 26.8224 m/s
This ensures that calculations involving speed are aligned with SI unit measurements, such as time in seconds or distance in meters. Converting units correctly is crucial for the success of solving physics problems.
Equations of Motion
The equations of motion allow us to calculate various components of an object's movement, assuming constant acceleration. They help to determine quantities like velocity, displacement, and time. The first equation of motion is:
  • \( v = u + at \)
Where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is acceleration
  • \( t \) is time
In our problem, the car starts from rest, so \( u = 0 \). If \( v = 26.8224 \) m/s and \( t = 10 \) s, we can solve for \( a \). The second equation of motion is:
  • \( s = ut + 0.5at^2 \)
Here, \( s \) represents the displacement, allowing us to calculate how far an object has traveled over time. Apply these equations step by step for calculating various aspects of motion under a constant acceleration.
Gravitational Force
Gravitational force is the attraction between two masses. On Earth, this force results in the acceleration known as gravity, denoted as \( g \), which is approximately 9.8 m/s虏. When comparing other accelerations to gravity, one may express acceleration as a fraction of \( g \). This offers perspective on how the force or acceleration compares to what we experience due to gravity daily.For instance, if an object has an acceleration of 2.68224 m/s虏, it can be compared to \( g \) by:
  • Calculating: \( a/g = 2.68224 \ m/s^2 / 9.8 \ m/s^2 \approx 0.274 \)
This means the object's acceleration is 0.274 times that of gravitational acceleration.
Distance Displacement
Distance and displacement describe an object's movement but are different concepts. Distance refers to the entire path traveled, while displacement measures the shortest distance from the start to end point. Using the equations of motion to find displacement involves:
  • Starting with: \( s = ut + 0.5at^2 \)
For a car starting from rest, and with a calculated acceleration, \( a \), over a time, \( t \), you find \( s \), the displacement. It's important for physics calculations to know exactly where and how far an object has moved, particularly in studies involving forces and motion.
Metric Conversion
Metric conversion allows us to switch between different units of measurement within the metric system or convert between metric and imperial units, such as meters to feet. Understanding these conversions is essential for accurately interpreting and converting physical quantities. For example, converting a displacement from meters to feet uses:
  • The conversion factor: 1 m = 3.281 ft
Using this:
  • 134.112 meters becomes: 134.112 m 脳 3.281 ft/m = 439.963 ft
Making this conversion helps to contextualize results in a unit that might be more familiar to those accustomed to imperial measurements.

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Most popular questions from this chapter

A rubber ball is shot straight up from the ground with speed \(v_{0}\) Simultaneously, a second rubber ball at height \(h\) directly above the first ball is dropped from rest. a. At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of \(v_{0}\) and \(g\). b. What is the maximum value of \(h\) for which a collision occurs before the first ball falls back to the ground? c. For what value of \(h\) does the collision occur at the instant when the first ball is at its highest point?

A particle's position on the \(x\) -axis is given by the function \(x=\left(t^{2}-4 t+2\right) \mathrm{m}\) where \(t\) is in \(\mathrm{s}\) a. Make a position-versus-time graph for the interval 0 s \(\leq\) \(t \leq 5\) s. Do this by calculating and plotting \(x\) every 0.5 s from \(0 \mathrm{s}\) to \(5 \mathrm{s}\), then drawing a smooth curve through the points. b. Determine the particle's velocity at \(t=1.0 \mathrm{s}\) by drawing the tangent line on your graph and measuring its slope. c. Determine the particle's velocity at \(t=1.0 \mathrm{s}\) by evaluating the derivative at that instant. Compare this to your result from part b. d. Are there any turning points in the particle's motion? If so, at what position or positions? e. Where is the particle when \(v_{x}=4.0 \mathrm{m} / \mathrm{s} ?\) f. Draw a motion diagram for the particle.

A car starts at the origin and moves with velocity \(\vec{v}=\) \((10 \mathrm{m} / \mathrm{s},\) northeast). How far from the origin will the car be after traveling for 45 s?

A driver has a reaction time of \(0.50 \mathrm{s}\), and the maximum deceleration of her car is \(6.0 \mathrm{m} / \mathrm{s}^{2} .\) She is driving at \(20 \mathrm{m} / \mathrm{s}\) when suddenly she sees an obstacle in the road \(50 \mathrm{m}\) in front of her. Can she stop the car in time to avoid a collision?

A car starts from rest at a stop sign. It accelerates at \(4.0 \mathrm{m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{s},\) coasts for \(2.0 \mathrm{s},\) and then slows down at a rate of \(3.0 \mathrm{m} / \mathrm{s}^{2}\) for the next stop sign. How far apart are the stop signs?
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