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Chapter 2: Problem 6

(II) You are driving home from school steadily at \(95 \mathrm{~km} / \mathrm{h}\) for \(130 \mathrm{~km}\). It then begins to rain and you slow to \(65 \mathrm{~km} / \mathrm{h}\). You arrive home after driving 3 hours and 20 minutes. (a) How far is your hometown from school? (b) What was your average speed?

Short Answer

Expert verified
(a) 257.73 km, (b) 77.32 km/h

Step by step solution

01

Convert total driving time to hours

The total driving time is 3 hours and 20 minutes. To convert this into hours, recognize that 20 minutes is \( \frac{20}{60} = \frac{1}{3} \) hour. Thus, the total time in hours is \( 3 + \frac{1}{3} = \frac{10}{3} \) hours.
02

Calculate the time driven at initial speed

First, calculate the time it takes to drive the initial 130 km at 95 km/h: \[ \text{Time} = \frac{130 \text{ km}}{95 \text{ km/h}} = \frac{130}{95} \text{ hours} = \frac{26}{19} \text{ hours} \approx 1.368 \text{ hours} \].
03

Determine the remaining driving time

Subtract the time driven at 95 km/h from the total driving time to find the remaining time: \[ \text{Remaining time} = \frac{10}{3} - \frac{26}{19} \approx 3.333 - 1.368 = 1.965 \text{ hours} \].
04

Calculate the distance for the second part of the trip

Using the remaining time and driving at 65 km/h, calculate the distance: \[ \text{Distance} = 65 \text{ km/h} \times 1.965 \text{ hours} = 127.73 \text{ km} \].
05

Find the total distance from school to hometown

Add the distances from both parts of the journey: \[ 130 \text{ km} + 127.73 \text{ km} = 257.73 \text{ km} \].
06

Determine the average speed for the whole journey

The average speed formula is \( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \). Substituting the known values gives: \[ \text{Average speed} = \frac{257.73 \text{ km}}{\frac{10}{3} \text{ hours}} = 77.32 \text{ km/h} \].

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Key Concepts

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

Average Speed
Average speed is a vital concept in kinematics that helps us understand how fast an object is moving over a period of time. It鈥檚 calculated using the formula:
\[\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}\]This means that to find the average speed of a trip, you need to know both the total distance traveled and the total time it took.
In the exercise, after calculating the total distance to be 257.73 km and the total time as \( \frac{10}{3} \approx 3.333 \) hours, you can find the average speed:
  • Plug in the total distance and time into the formula.
  • This gave us 77.32 km/h for the entire journey.
This tells us how fast the journey was on average, considering both the fast drive in clear weather and the slower speed in the rain.
Distance Calculation
Distance calculation involves determining how far an object has traveled. It is done by multiplying speed by time:
\[\text{Distance} = \text{Speed} \times \text{Time}\]In the exercise, we had to determine the distance of both parts of the journey:
  • First part: the car traveled 130 km at 95 km/h.
  • Second part: the car kept moving at a lower speed of 65 km/h to cover the remaining distance.
For the second leg of the trip, after figuring out the remaining time was approximately 1.965 hours:
  • The distance was computed by multiplying 65 km/h by 1.965 hours, resulting in approximately 127.73 km.
These calculations are crucial as they help us figure out the total journey distance, which is essential for finding the average speed.
Time Conversion
Time conversion is a basic yet critical skill in kinematics, often necessary to ensure that units align correctly in calculations.
In this problem, we started with a time of 3 hours and 20 minutes. To work with this conveniently, we convert minutes into hours.

Knowing that 1 hour equals 60 minutes:
  • 20 minutes is converted as \( \frac{20}{60} = \frac{1}{3} \) of an hour.
  • This converts the total travel time to \( 3 + \frac{1}{3} = \frac{10}{3} \) or approximately 3.333 hours.
Clearly understanding this conversion ensures that calculations are performed with consistent units, which is absolutely essential for correctly applying kinematic formulas like those for speed and distance.

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Most popular questions from this chapter

(I) A car accelerates from \(12 \mathrm{~m} / \mathrm{s}\) to \(21 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~s}\). What was its acceleration? How far did it travel in this time? Assume constant acceleration.

(II) A baseball is seen to pass upward by a window \(23 \mathrm{~m}\) above the street with a vertical speed of \(14 \mathrm{~m} / \mathrm{s}\). If the ball was thrown from the street, \((a)\) what was its initial speed, (b) what altitude does it reach, \((c)\) when was it thrown, and ( \(d\) ) when does it reach the street again?

(II) Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius \(R_{1}=2.5 \mathrm{~cm}\) and finishes at radius \(R_{2}=5.8 \mathrm{~cm} .\) The distance between the centers of neighboring spiralwindings is \(1.6 \mu \mathrm{m}\left(=1.6 \times 10^{-6} \mathrm{~m}\right) .\) (a) Determine the total length of the spiraling path. [Hint: Imagine "unwinding" the spiral into a straight path of width \(1.6 \mu \mathrm{m}\), and note that the original spiral and the straight path both occupy the same area.] (b) To read information, a CD player adjusts the rotation of the CD so that the player's readout laser moves along the spiral path at a constant speed of \(1.25 \mathrm{~m} / \mathrm{s} .\) Estimate the maximum playing time of such a CD.

(III) The acceleration of a particle is given by \(a=A \sqrt{t}\) where \(A=2.0 \mathrm{~m} / \mathrm{s}^{5 / 2} .\) At \(t=0, v=7.5 \mathrm{~m} / \mathrm{s}\) and \(x=0\) (a) What is the speed as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, speed and displacement at \(t=5.0 \mathrm{~s} ?\)

(II) An inattentive driver is traveling \(18.0 \mathrm{~m} / \mathrm{s}\) when he notices a red light ahead. His car is capable of decelerating at a rate of \(3.65 \mathrm{~m} / \mathrm{s}^{2} .\) If it takes him \(0.200 \mathrm{~s}\) to get the brakes on and he is \(20.0 \mathrm{~m}\) from the intersection when he sees the light, will he be able to stop in time?
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