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You jog at \(9.50 \mathrm{~km} / \mathrm{h}\) for \(8.00 \mathrm{~km}\); then you jump into a car and ride an additional \(16.0 \mathrm{~km}\). What average speed must the car have for the average speed for the entire \(24.0-\mathrm{km}\) trip to be \(22.0 \mathrm{~km} / \mathrm{h}\) ?

Short Answer

Expert verified
The car must average about 64.3 km/h.

Step by step solution

01

Determine Total Distance and Desired Average Speed

The total distance of the trip is given as \(24.0\, \text{km}\). The desired average speed for the trip is \(22.0\, \text{km/h}\). We need to calculate the average speed of the car for the entire trip to result in this average speed.
02

Calculate Total Time Needed for the Desired Average Speed

For the total trip of \(24.0\, \text{km}\) to have an average speed of \(22.0\, \text{km/h}\), we use the formula: \(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\). Thus, the total time \(T\) needed can be calculated as: \[ T = \frac{24.0}{22.0} \approx 1.0909 \, \text{hours} \]
03

Calculate Time Spent Jogging

The jogging phase covers \(8.0\, \text{km}\) at a speed of \(9.50\, \text{km/h}\). The time spent jogging \(t_1\) can be calculated with the formula: \[ t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{8.0}{9.50} \approx 0.8421 \, \text{hours} \]
04

Determine Time Available for Car Ride

Subtract the jogging time from the total time to find the time available for the car ride \(t_2\): \[ t_2 = T - t_1 \approx 1.0909 - 0.8421 = 0.2488 \, \text{hours} \]
05

Calculate Required Car Speed

To find the necessary speed of the car \(v_c\), use the formula \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\). The car must cover \(16.0\, \text{km}\) in approximately \(0.2488\, \text{hours}\). Hence: \[ v_c = \frac{16.0}{0.2488} \approx 64.3 \, \text{km/h} \]

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Key Concepts

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

Distance Calculation
When it comes to finding out how far you need to travel, calculating distance is quite useful. In our exercise, you work out two main parts of the journey: jogging and driving in a car. The total distance for both is given as 24 km. You jogged 8 km first, followed by a 16 km car ride.

These calculations rely on simple addition:
  • Jogged distance = 8 km
  • Car ride distance = 16 km
  • Total journey distance = 8 km + 16 km = 24 km
Once you have the components of your journey, knowing how far you'll travel helps in planning overall speed and time.
Speed and Velocity
Speed is just how fast you go, while velocity adds direction to speed. In the context of this exercise, we focus on speed to determine how quickly you complete your trip. The target is an average speed, which makes it interesting!

Average speed is key because it reflects your overall pace rather than individual parts. To calculate average speed:
  • Use the formula: \( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \)
  • For 24 km at 22 km/h, you aim for a consistent pace across all segments.
This concept helps you assess how much faster you need to drive relative to jogging to meet the trip's speed goal. Understanding speed and velocity ensures efficient travel and helps set realistic journey expectations.
Time Calculation
Time calculation is vital for managing how long each part of your trip takes. By breaking down the journey into segments, you estimate the jogging and car times efficiently. Here’s how this works in the exercise:
  • Jogging time \( t_1 = \frac{8.0\, \text{km}}{9.50\, \text{km/h}} \approx 0.8421\, \text{hours} \)
  • Total time desired for entire trip \( T = \frac{24.0\, \text{km}}{22.0\, \text{km/h}} \approx 1.0909\, \text{hours} \)
  • Time for car drive \( t_2 = T - t_1 \approx 1.0909 - 0.8421 = 0.2488 \text{hours} \)
From these calculations, knowing the time ensures balanced timing across activities. It helps figure out how quick the car trip needs to be to meet the target total average speed. Proper time management allows for realistic travel planning, using each segment wisely.

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Most popular questions from this chapter

Predict \& Explain You drive your car in a straight line at \(15 \mathrm{~m} / \mathrm{s}\) for \(10 \mathrm{~min}\), then at \(25 \mathrm{~m} / \mathrm{s}\) for another \(10 \mathrm{~min}\). (a) Is your average speed for the entire trip more than, less than, or equal to \(20 \mathrm{~m} / \mathrm{s}\) ? (b) Choose the best explanation from the following: A. More time is required to drive the same distance at \(15 \mathrm{~m} / \mathrm{s}\) than at \(25 \mathrm{~m} / \mathrm{s}\). B. Less distance is covered driving at \(25 \mathrm{~m} / \mathrm{s}\) than at \(15 \mathrm{~m} / \mathrm{s}\). C. Equal time is spent driving at \(15 \mathrm{~m} / \mathrm{s}\) and \(25 \mathrm{~m} / \mathrm{s}\).

Follow-up Suppose the kingfisher dives with an average speed of \(4.6 \mathrm{~m} / \mathrm{s}\) for \(1.4 \mathrm{~s}\) before hitting the water. What was the height from which the bird dove?

Triple Choice An object's position-time graph is a straight line with a positive slope. Is the velocity of this object positive, negative, or zero? Explain.

Can you take a walk in such a way that the distance you cover is greater than the magnitude of your displacement? Give an example if your answer is yes; explain why not if your answer is no.

Think \& Calculate A train travels in a straight line at \(20.0 \mathrm{~m} / \mathrm{s}\) for \(2 \mathrm{~km}\), then at \(30.0 \mathrm{~m} / \mathrm{s}\) for another \(2 \mathrm{~km}\). (a) Is the average speed of the train greater than, less than, or equal to \(25 \mathrm{~m} / \mathrm{s}\) ? Explain. (b) Verify your answer to part (a) by calculating the average speed.

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