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Chapter 2: Problem 83

A golfer rides in a golf cart at an average speed of \(3.10 \mathrm{m} / \mathrm{s}\) for \(28.0 \mathrm{s} .\) She then gets out of the cart and starts walking at an average speed of \(1.30 \mathrm{m} / \mathrm{s} .\) For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is \(1.80 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
The golfer must walk for 72.8 seconds.

Step by step solution

01

Determine the distance covered while riding

To find the distance covered while riding, use the formula for distance: \( \text{distance} = \text{speed} \times \text{time} \). Here, the speed while riding is \(3.10\, \mathrm{m/s}\) and the time is \(28.0\, \mathrm{s}\). Thus, the distance is \(3.10\, \mathrm{m/s} \times 28.0\, \mathrm{s} = 86.8\, \mathrm{m}\).
02

Express the walking distance in terms of time

Let the time spent walking be \( t_w \) seconds. The distance covered while walking is then given by \( \text{speed} \times \text{time} = 1.30\, \mathrm{m/s} \times t_w \).
03

Write the equation for average speed of the trip

The average speed for the entire trip which includes both riding and walking is given by the total distance divided by total time. Total distance \( = 86.8 \text{ m} + 1.30t_w \text{ m} \) and total time \( = 28.0 \text{ s} + t_w \text{ s} \). Average speed is given: \[ 1.80 = \frac{86.8 + 1.30t_w}{28.0 + t_w} \].
04

Solve for the walking time

Cross-multiply to clear the fraction and solve the equation: 1.80\( \times (28.0 + t_w) = 86.8 + 1.30t_w \).2. This simplifies to: \[ 1.80 \times 28.0 + 1.80t_w = 86.8 + 1.30t_w \] \[ 50.4 + 1.80t_w = 86.8 + 1.30t_w \] 3. Rearrange terms to find \(t_w\): \[ 1.80t_w - 1.30t_w = 86.8 - 50.4 \] 4. Simplify to solve for \(t_w\): \[ 0.50t_w = 36.4 \] 5. Divide by 0.50: \[ t_w = \frac{36.4}{0.50} = 72.8 \]
05

Conclusion

The golfer must walk for \(72.8\, \text{s}\) to achieve an average speed of \(1.80\, \text{m/s}\) for the entire trip.

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Key Concepts

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

Distance Calculation
To understand distance calculation, think about the relationship between speed, time, and distance. In physics, distance is calculated by multiplying speed (how fast something is moving) by time (how long it's moving).
For example, if you ride a golf cart at a speed of 3.10 meters per second for 28 seconds, the distance covered is calculated as follows:
  • Use the formula:
    \( \text{Distance} = \text{Speed} \times \text{Time} \)
  • Plug in the values:
    \( \text{Distance} = 3.10 \, \text{m/s} \times 28.0 \, \text{s} \)
  • The result is:
    \( \text{Distance} = 86.8 \, \text{m} \)
This method helps us find out how far the golfer travels during the cart ride.
Speed and Time Relationship
Speed and time have a direct relationship in calculating both distance and average speed. When you know two of these variables, you can easily find the third one. Understanding this relationship is key to solving many physics problems.
In the problem where the golfer walks at a different speed after riding a cart, we aim to find the time it takes for her to reach a required average speed for the entire journey.
The time spent walking needs to satisfy:
  • Total distance involves both cart and walking distances:
    \(86.8 + 1.30 \, \mathrm{m/s} \times t_w\)
  • Total time is the cart time plus walking time:
    \(28.0 + t_w\)
  • Average speed requested:
    \( \frac{86.8 + 1.30t_w}{28.0 + t_w} = 1.80 \, \mathrm{m/s} \)
Understanding this relationship allows us to calculate the necessary walking time.
Physics Problem Solving
Physics problems often require step-by-step solutions like the one we solved for the golfer. These problems build your ability to approach real-world situations with logical reasoning.
Key steps include:
  • Identifying the known values and what is required.
  • Sketching out what equations relate these values (distance = speed \(\times\) time, average speed formula).
  • Substituting the known values into these equations.
  • Simplifying the equations (such as clearing fractions) to isolate the variable (in our case, \(t_w\)).
By cross-referencing the information provided and using the relevant formulas, these problems become more manageable and less prone to mistakes. It's all about establishing a clear sequence in tackling the problem.
Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces causing the motion. It primarily focuses on the relationship between displacement, velocity (speed in this context), and acceleration over time.
In the given problem involving the golfer, the concept of kinematics helps us reference:
  • The journey, which shows a change from one motion (riding in a cart) to another (walking).
  • Understanding both modes of motion using their speeds and the time involved without delving into why the speeds are what they are.
  • Calculating the result (average speed for the trip) which is a basic application of kinematic principles without complicating factors like external forces.
Kinematics provides the core tools and language we use to describe these kinds of physical scenarios, simplifying what could otherwise be complex calculations into more digestible steps.

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Most popular questions from this chapter

A car is traveling at a constant speed of \(33 \mathrm{m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{km}\) away?

A locomotive is accelerating at \(1.6 \mathrm{m} / \mathrm{s}^{2} .\) It passes through a \(20.0-\mathrm{m}-\) wide erossing in a time of \(2.4 \mathrm{s}\). After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{m} / \mathrm{s} ?\)

A runner is at the position \(x=0 \mathrm{m}\) when time \(t=0 \mathrm{s}\). One hundred meters away is the finish line. Every ten seconds, this runner runs half the remaining distance to the finish line. During each ten-second segment, the runner has a constant velocity. For the first forty seconds of the motion, construct (a) the position-time graph and (b) the velocity-time graph.

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of \(86.0 \mathrm{m} / \mathrm{s}^{2}\) for 1.70 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

A jet is taking off from the deck of an aircraft carrier, as shown in the image. Starting from rest, the jet is catapulted with a constant acceleration of \(+31 \mathrm{m} / \mathrm{s}^{2}\) along a straight line and reaches a velocity of \(+62 \mathrm{m} / \mathrm{s} .\) Find the displacement of the jet.
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