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You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65 mi/h), and the trip takes 1 h and 50 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 km/h (43 mi/h). How much longer does the trip take?

Short Answer

Expert verified
The trip takes 55 minutes longer on Friday afternoon.

Step by step solution

01

Convert Trip Time to Hours

The trip normally takes 1 hour and 50 minutes. To convert this into hours, consider that 50 minutes is 50/60 hours. So, the time is: \[ 1 + \frac{50}{60} = 1.833\text{ hours.} \]
02

Calculate Normal Trip Distance

Use the formula for distance, \( \text{Distance} = \text{Speed} \times \text{Time} \). Given the speed is 105 km/h and the time is 1.833 hours, compute: \[ 105 \text{ km/h} \times 1.833 \text{ hours} = 192.465\text{ km.} \]
03

Calculate Time for Slower Speed

When driving at 70 km/h, use the distance calculated in Step 2. Use the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). Compute the time: \[ \frac{192.465 \text{ km}}{70 \text{ km/h}} = 2.75\text{ hours.} \]
04

Calculate Extra Time

To find the additional travel time, subtract the normal time from the time taken with the slower speed: \[ 2.75\text{ hours} - 1.833\text{ hours} = 0.917\text{ hours.} \]
05

Convert Extra Time to Minutes

Convert the extra time back to minutes since 0.917 hours need to be expressed in minutes: \[ 0.917 \times 60 = 55\text{ minutes.} \]

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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 key concept in understanding how quickly or slowly a journey takes place. It is defined as the total distance traveled divided by the total time taken. In simple terms, it is how fast you are moving on average over a stretch of your journey.
For example, when driving between San Diego and Los Angeles at an average speed of 105 km/h, your average speed indicates that for every hour you travel, you cover 105 kilometers.
To further simplify:
  • Average speed helps in determining your overall rate of motion.
  • It smoothens out all variations in speed that might occur during the trip.
  • It's crucial for calculating travel time and planning journeys effectively.
Average speed can hugely affect how long a trip takes, just like in the case of heavy traffic decreasing the speed to 70 km/h and thereby increasing travel time.
Distance Calculation
Distance calculation involves determining the length of the path traveled during a journey. It is an essential aspect as it directly impacts the time required to make a trip.
The formula to calculate distance is: \[ \text{Distance} = \text{Speed} \times \text{Time} \] In the example where you travel from San Diego to Los Angeles under normal conditions, the distance was calculated using an average speed of 105 km/h over 1.833 hours. This gives:
  • \[ 105 \text{ km/h} \times 1.833 \text{ hours} = 192.465\text{ km} \]
Understanding distance calculation is crucial because it lays the foundation for computing other travel parameters, such as estimated travel time at various speeds.
Time Conversion
Time conversion is about changing time from one unit to another and is often necessary for calculations involving different units of time. For example, you may need to convert minutes into hours or vice versa to align with speed units like km/h.
In exercises like the freeway trip from San Diego to Los Angeles, time conversion takes place when converting 1 hour and 50 minutes into purely hours by doing:
  • Convert 50 minutes into hours as: \[ \frac{50}{60} \text{ hours} = 0.833 \text{ hours} \]
  • Added to the initial 1 hour, gives: \[ 1 + 0.833 = 1.833 \text{ hours} \]
This process allows for more straightforward calculations when matching against speeds represented in the same units (hours in this case). It's crucial in creating cohesive and correct calculations.
Travel Time Calculation
Travel time calculation allows you to estimate how long a journey will take given a particular speed and distance. Calculating travel time is necessary for planning and adjusting your expectations for travel durations under different conditions.
You use the formula:
  • \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
For instance, when calculating how long it takes to travel 192.465 km at a reduced speed of 70 km/h, you plug into the formula to get:
  • \[ \frac{192.465 \text{ km}}{70 \text{ km/h}} = 2.75 \text{ hours} \]
Recognizing how traffic conditions affect speed, and thus travel time can lead to added time on your journey, such as the extra 0.917 hours calculated, which converts to about 55 minutes. Understanding these principles enhances planning and accurate time management.

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Most popular questions from this chapter

A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by \(x(t) = bt^2 - ct^3\), where \(b =\) 2.40 m/s\(^2\) and \(c =\) 0.120 m/s\(^3\). (a) Calculate the average velocity of the car for the time interval \(t =\) 0 to \(t =\) 10.0 s. (b) Calculate the instantaneous velocity of the car at \(t =\) 0, \(t =\) 5.0 s, and \(t =\) 10.0 s. (c) How long after starting from rest is the car again at rest?

A small object moves along the \(x\) -axis with acceleration \(a_{x}(t)=-\left(0.0320 \mathrm{~m} / \mathrm{s}^{3}\right)(15.0 \mathrm{~s}-t) .\) At \(t=0\) the object is at \(x=-14.0 \mathrm{~m}\) and has velocity \(v_{0 x}=8.00 \mathrm{~m} / \mathrm{s} .\) What is the \(x\) -coordinate of the object when \(t=10.0 \mathrm{~s} ?\)

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs for the motion.

The rocket-driven sled \(\textit{Sonic Wind No. 2,}\) used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s(500 mi/h) in 0.900 s. (a) Compute the acceleration in m/s\(^2\), assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body (\(g\))? (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled decreased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40\(g\). Are these figures consistent?

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{~s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{~s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

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