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Chapter 2: Problem 3

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

Short Answer

Expert verified
Thus, with the reduced speed due to traffic, the trip would take approximately 55 minutes longer than usual.

Step by step solution

01

Calculate the Distance

To calculate the distance from San Diego to Los Angeles, we multiply the speed during normal driving conditions by the time. The speed is \(105 \, \mathrm{km/h}\) and the time taken is 1 hour 50 minutes which we convert to hours for consistency, that is \(1.833 \, \mathrm{hours}\). Hence, the distance \(d\) can be calculated by the formula \( d = speed \times time = 105 \, \mathrm{km/h} \times 1.833 \, \mathrm{hours} = 192.465 \, \mathrm{km}\)
02

Calculate the New Time

Now we want to find how long it takes to travel this distance at a slower speed of 70 km/h. We rearrange the same formula we used above to find time. The new time (\(t')\), will be the distance divided by the slower speed. So, \(t' = d \div speed = 192.465 \, \mathrm{km} \div 70 \, \mathrm{km/h} = 2.7495 \, \mathrm{hours}\) which is approximately 2 hours and 45 minutes.
03

Calculate the Additional Time

We then deduct the original time from the new time to find out how much longer does the trip take due to the slowed down traffic. The extra time will hence be \(t' - t = 2.7495 \, \mathrm{hours} - 1.833 \, \mathrm{hours} = 0.916 \, \mathrm{hours}\), which is approximately 55 minutes. So the trip will take about 55 minutes longer.

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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 measure of the total distance traveled divided by the total time taken to cover that distance. It's a way to quantify how fast something is moving overall, and is usually expressed in units like kilometers per hour (km/h) or miles per hour (mi/h).

For our freeway trip from San Diego to Los Angeles, if you normally travel at an average speed of 105 km/h for 1 hour and 50 minutes, you're making pretty good time! However, if heavy traffic reduces your average speed to 70 km/h, you'll spend much more time on the road. Despite this change in speed, the distance remains the same, but the time taken to traverse this distance increases, leading to a longer trip duration.
Distance Calculation
Distance calculation is key in planning trips and understanding travel times. To find the distance, you multiply the speed at which you're traveling by the time you spend traveling. This is useful in many real-world situations like planning travel times or finding out how far you've walked or driven.

In the exercise, by traveling at 105 km/h for 1 hour and 50 minutes, the distance is calculated by converting the time into hours (1.833 hours) and then multiplying it by the speed, resulting in a distance of approximately 192.465 km between San Diego and Los Angeles. Knowing the distance is crucial as it remains constant regardless of the speed and is essential in calculating the new time needed to travel at a different speed.
Time Conversion
Time conversion is a useful skill that involves converting units of time to each other, such as minutes to hours or vice versa. It's necessary when dealing with calculations in physics, especially in kinematics, like when you are trying to determine how long a trip will take.

In the original problem, you're given a time in hours and minutes, which is then converted into just hours to maintain consistency with the speed given in km/h. For example, 1 hour and 50 minutes is converted to 1.833 hours, which is then used in the distance calculation. It's essential to convert time correctly, as failing to do so can drastically affect the results of calculations, such as the time it takes to complete a trip.
Velocity-Time Relationship
The velocity-time relationship is fundamental in kinematics and refers to how the velocity of an object changes over time. Velocity, which incorporates speed and direction, directly affects how much time it will take for an object to travel a certain distance.

In instances like the textbook exercise, when the average speed decreases due to heavy traffic, there's a direct impact on travel time: the distance remains the same but because the velocity (speed) has decreased, it will take a longer time to cover that distance. This relationship is crucial in calculating the difference in trip duration, and understanding it helps in various applications, from driving to physics experiments.

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Most popular questions from this chapter

If the contraction of the left ventricle lasts \(250 \mathrm{~ms}\) and the speed of blood flow in the aorta (the large artery leaving the heart) is \(0.80 \mathrm{~m} / \mathrm{s}\) at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? (a) \(310 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{b}) 31 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{c}) 3.2 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{d}) 0.32 \mathrm{~m} / \mathrm{s}^{2} .\)

A student is running at her top speed of \(5.0 \mathrm{~m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still \(40.0 \mathrm{~m}\) from the bus, it starts to pull away, moving with a constant acceleration of \(0.170 \mathrm{~m} / \mathrm{s}^{2}\). (a) For how much time and what distance does the student have to run at \(5.0 \mathrm{~m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is \(3.5 \mathrm{~m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

A turtle crawls along a straight line, which we'll call the \(x\) -axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{~cm}+\) \((2.00 \mathrm{~cm} / \mathrm{s}) t-\left(0.0625 \mathrm{~cm} / \mathrm{s}^{2}\right) t^{2} .\) (a) Find the turtle's initial velocity. initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of \(10.0 \mathrm{~cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of \(x\) versus \(t, v_{1}\) versus \(t,\) and \(a_{f}\) versus \(t,\) for the time interval \(t=0\) to \(t=40 \mathrm{~s}\)

A rock is thrown straight up with an initial speed of \(24.0 \mathrm{~m} / \mathrm{s}\) Neglect air resistance. (a) At \(t=1.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing? (b) At \(t=3.0 \mathrm{~s}\), what are the directions of the velocity and acceleration of the rock? Is the speed of the rock increasing or decreasing?

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes \(10 \mathrm{~s}\) to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H,\) how high (in terms of \(H\) ) will the faster stone go? Assume free fall.
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