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Chapter 2: Problem 10

To set a speed record in a measured (straight-line) distance \(d\), a race car must be driven first in one direction (in time \(t_{1}\) ) and then in the opposite direction (in time \(t_{2}\) ). (a) To eliminate the effects of the wind and obtain the car's speed \(v_{c}\) in a windless situation, should we find the average of \(d / t_{1}\) and \(d / t_{2}\) (method 1 ) or should we divide \(d\) by the average of \(t_{1}\) and \(t_{2}\) ? (b) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed \(v_{w}\) to the car's speed \(v_{c}\) is \(0.0240\) ?

Short Answer

Expert verified
Use Method 2 for accuracy: divide distance by average time. Fractional difference is 0.000576.

Step by step solution

01

Understanding the Problem

The goal is to find the best method to measure the car's speed in a windless situation by averaging the speeds from two trials in opposite directions. You need to determine which averaging method more accurately eliminates the effects of the wind.
02

Analyzing Method 1

For method 1, we calculate the speeds of the car going in both directions as \(v_{1} = \frac{d}{t_{1}}\) and \(v_{2} = \frac{d}{t_{2}}\). Then, we compute the average speed as \(\bar{v}_{1} = \frac{1}{2}\left(\frac{d}{t_{1}} + \frac{d}{t_{2}}\right)\).
03

Analyzing Method 2

For method 2, compute the average time as \(\bar{t} = \frac{1}{2}(t_{1} + t_{2})\). Next, find the average speed by dividing the distance by this average time: \(\bar{v}_{2} = \frac{d}{\bar{t}} = \frac{2d}{t_{1} + t_{2}}\).
04

Evaluating Method 1 with Wind Effects

Consider the speed of the car with and against the wind: \(v_{1} = v_{c} + v_{w}\) and \(v_{2} = v_{c} - v_{w}\). Then, \( \bar{v}_{1} = \frac{1}{2}\left( \frac{d}{t_{1}} + \frac{d}{t_{2}} \right) = \frac{v_{1} + v_{2}}{2} = v_{c}\) since \(v_{1} = \frac{d}{t_{1}} = v_{c} + v_{w}\) and \(v_{2} = \frac{d}{t_{2}} = v_{c} - v_{w}\). The wind effects are not entirely canceled.
05

Evaluating Method 2 with Wind Effects

For method 2, we evaluate the average speed calculation: \(\bar{t} = \frac{1}{2}(t_{1} + t_{2})\) and \(\bar{v}_{2} = \frac{d}{\bar{t}} = \frac{2d}{t_{1} + t_{2}}\). Substituting \(t_{1} = \frac{d}{v_{c}+v_{w}}\) and \(t_{2} = \frac{d}{v_{c}-v_{w}}\), the wind effects cancel and \(\bar{v}_{2} = v_{c}\).
06

Comparing Fractional Difference

To find the fractional difference, the formula is \(\frac{\bar{v}_{1} - \bar{v}_{2}}{\bar{v}_{2}}\). Given \(v_{w}/v_{c} = 0.0240\), calculate \[\frac{\bar{v}_{1} - \bar{v}_{2}}{\bar{v}_{2}} = \frac{\left(\frac{v_{c}^2 - v_{w}^2}{v_{c}}\right) - v_{c}}{v_{c}} \approx 0.0240^2\]. New estimate \(\approx 0.000576\).
07

Conclusion and Select Method

Method 2 is more accurate as it effectively cancels out the wind effects. The fractional difference is approximately \(0.000576\) when the wind affects the average speed.

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Key Concepts

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

Speed Measurement
Measuring speed involves calculating how fast an object moves over a certain distance in a given time. To calculate the speed of a car over a straight-line distance, use the formula:
  • \( v = \frac{d}{t} \)
Here:
  • \( v \) is the speed of the car,
  • \( d \) is the distance traveled, and
  • \( t \) is the time taken to travel that distance.
Speed measurement is crucial in many applications, like in transportation and physics experiments. Reliable speed measurements require consistent tests under standardized conditions. When measuring speed for a record, you ensure that all external factors, such as wind, don't skew results significantly. To combat this, average speed calculations are often employed, taking multiple measurements and using different methods to find an accurate result.
Wind Effects
Wind can either boost or hinder an object's movement, affecting its speed reading. A tailwind (wind moving with the object) can increase speed, while a headwind (against the object) decreases it. This influence means that one direction may result in a faster speed reading compared to the other when traveling the same route. In speed measurements, wind impacts must be carefully managed to obtain true speed values. One way to handle this is by conducting tests in both directions over the same distance. This approach helps ensure that wind effects, pushing the object in one direction and resisting it in the opposite, are counteracted.

Neutralizing Wind Effects

By measuring the time taken to travel a distance both downwind and upwind, wind effects can become negligible. When you calculate the average using both measurements, you better arrive at the speed that would be observed in still air. Method 2 from our exercise is helpful here; it calculates the speed using the total distance and average time, thus effectively canceling the wind's influence.
Average Speed Calculation
Calculating average speed involves a comprehensive look at multiple speed measures over the same distance. It offers a more reliable estimate, especially in conditions where external factors like wind could affect individual measurements.For example, if a car travels a distance \( d \) in one direction with time \( t_1 \) and returns in time \( t_2 \), two methods can be used to calculate average speed:
  • Method 1: Taking the arithmetic mean of speeds in both directions: \( \bar{v}_1 = \frac{1}{2} \left( \frac{d}{t_1} + \frac{d}{t_2} \right) \).
  • Method 2: Dividing the total distance by the average of the travel times: \( \bar{v}_2 = \frac{2d}{t_1 + t_2} \).

Which Method is Better?

While Method 1 simply averages the calculated speeds, Method 2 uses the total travel time to find a more representative speed unaffected by one-off variations like wind. As shown in the exercise, Method 2 more accurately reflects the car鈥檚 speed under windless conditions, nullifying wind effects and providing a consistent and valid measurement.

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Most popular questions from this chapter

You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of \(v_{0}=55 \mathrm{~km} / \mathrm{h}\); your best deceleration rate has the magnitude \(a=5.18 \mathrm{~m} / \mathrm{s}^{2}\). Your best reaction time to begin braking is \(T=0.75 \mathrm{~s}\). To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at \(55 \mathrm{~km} / \mathrm{h}\) if the distance to the intersection and the duration of the yellow light are (a) \(40 \mathrm{~m}\) and \(2.8 \mathrm{~s}\), and (b) \(32 \mathrm{~m}\) and \(1.8 \mathrm{~s}\) ? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).

Catapulting mushrooms. Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop's weight, but when the film reaches the drop, the drop's water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of \(1.6 \mathrm{~m} / \mathrm{s}\) in a \(5.0 \mu \mathrm{m}\) launch; its speed is then reduced to zero in \(1.0 \mathrm{~mm}\) by the air. Using those data and assuming constant accelerations, find the acceleration in terms of \(g\) during (a) the launch and (b) the speed reduction.

In \(1 \mathrm{~km}\) races, runner 1 on track 1 (with time \(2 \mathrm{~min}, 27.95 \mathrm{~s}\) ) appears to be faster than runner 2 on track 2 ( \(2 \mathrm{~min}, 28.15 \mathrm{~s}\) ). However, length \(L_{2}\) of track 2 might be slightly greater than length \(L_{1}\) of track 1. How large can \(L_{2}-L_{1}\) be for us still to conclude that runner 1 is faster?

When a high-speed passenger train traveling at \(161 \mathrm{~km} / \mathrm{h}\) rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance \(D=676 \mathrm{~m}\) ahead (Fig. 2-20). The locomotive is moving at \(29.0 \mathrm{~km} / \mathrm{h}\). The engineer of the high-speed train immediately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at \(x=0\) when, at \(t=0\), he first spots the locomotive. Sketch \(x(t)\) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.

A rocket, initially at rest, is fired vertically with an upward acceleration of \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). At an altitude of \(0.500 \mathrm{~km}\), the engine of the rocket cuts off. What is the maximum altitude it achieves?
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