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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

Sam heaves a 16 lb shot straight up, giving it a constant upward acceleration from rest of \(35.0 \mathrm{~m} / \mathrm{s}^{2}\) for \(64.0 \mathrm{~cm}\). He releases it \(2.20 \mathrm{~m}\) above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, \(1.83 \mathrm{~m}\) above the ground?

During an auto accident, the vehicle's airbags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, airbags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g}\) ?

A gazelle is running in a straight line (the \(x\) -axis). The graph in at a rate of \(1.60 \mathrm{~m} / \mathrm{s}^{2}\) for \(14.0 \mathrm{~s}\). It runs at constant speed for \(70.0 \mathrm{~s}\) and slows down at a rate of \(3.50 \mathrm{~m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for \(10.0 \mathrm{~s}\), Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back \(7.0 \mathrm{~s}\) after leaving the helicopter, and then he has a constant downward acceleration with magnitude \(2.0 \mathrm{~m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

The fastest measured pitched baseball left the pitcher's hand at a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). If the pitcher was in contact with the ball over a distance of \(1.50 \mathrm{~m}\) and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?
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