Physics

course Phy 201

6/1 9:45

Question: `q001. There are two parts to this problem. Reason them out using common sense.If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark?

Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?

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Your solution:

You need to move 10 mph. 10 mph / 2 mph per second = 5 seconds. It will take 5 seconds for the speedometer to change from 20 to 30 mph.

In 7 seconds the speedometer will move 7 seconds*2mph or 14 mph. Starting at 10 mph, after 7 seconds the meter will read 10+14 or 24 mph.

confidence rating #$&* 3

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Given Solution:

`aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph

Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph.

Will the vehicle require more or less than 10 seconds to reach the lamppost?

Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?

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Your solution:

The vehicle will require less time to reach the lamppost since it is going faster, it will cover the distance faster than the first time (less than 10 seconds). No, the speed will not be 10 mph greater than the first run because that measurement was based on time. Because during the second test less time is needed to get to the lamppost, the car will not pick up as much speed so it will not be 10 mph greater than before.

confidence rating #$&* 3

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Given Solution:

`aIf it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds.

The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second.

We wish to compare the rates at which two different automobiles increase their speed:

Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?

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Your solution:

Car A: 10 mph/5 sec = 2 mph/sec Car B: 50 mph/ 20 sec = 5/2 mph/sec

The rate of change is greater for the second car.

The car going from 40 mph to 90 mph in 20 seconds is speeding up at a greater rate.

confidence rating #$&* 3

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Given Solution:

The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of change in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second.

The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second.

Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.

Self-critique: OK

Self-critique rating #$&* OK

Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile.

Which team will win and why?

If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?

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Your solution:

Team A: 3000 Newtons/1500 kg = 2 N/kg Team B: 5000 N/2000 kg = 5/2 N/kg

The second team will win because their rate of change for movement is greater and will allow them to get to 5 mph quicker than the first team.

5000 N -500N/ 2000 kg = 4500 N/2000kg = 9/4 N/kg. With the 500 N resistance, the first team would still not win because the rate of change of the second team is sill greater than that of the first team.

confidence rating #$&* 3

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Given Solution:

`aThe first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead.

The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you predict will be moving backward immediately after the collision, and why?

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Your solution:

Player A: 250 lb*10fps = 2500 lbs fps Player B: 200 lb*20fps = 4000 lbs fps

The first player will be moving backward after the collision because the second player his him at a higher velocity. It canceled out the first player’s movement and there was extra force to cause him to move backward.

confidence rating #$&* 2

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Given Solution:

`aGreater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion.

It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision.

In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?

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Your solution:

Climber A: 200 lb / 12 oz = 16.67 Climber B: 150 lb / 10 oz = 15

The first climber should be able to climb farther because he has more energy.

confidence rating #$&* 1

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Given Solution:

`aThe comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight.

It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.

Self-critique (if necessary): I had the formula set up wrong. I should have had the ounces on the top of the fraction to get oz/lb for the energy. When you use the right formula, you find the climber with the lower oz/lb to be able to climb higher, which would be the second climber. This is because the second person has a higher energy to body weight ratio.

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Self-critique rating #$&* 3

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Question: `q007. Two automobiles are traveling up a long hill with a steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop.

Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long?

Which automobile will have the greater average coasting velocity? Will its average coasting velocity be twice as great as the other, more than twice as great or less than twice as great?

Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?

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Your solution:

The car that was going twice as fast as the other will take longer to stop because it has more energy that has to be released. It will take about twice as long for the faster car to stop than the slower one.

The car going twice as fast will have the greater average coasting velocity. The velocity will be about twice that of the slower car.

The distance travelled by the faster car will be twice that of the distance that the slower car travelled.

confidence rating #$&* 1

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Given Solution:

`aIt turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long.

If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower.

For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected.

Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far.

If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.

STUDENT COMMENT: I do not understand why the car would go four times as far as the slower car.

INSTRUCTOR RESPONSE: The faster car takes twice as long to come to rest, and have twice the average velocity.

If the car traveled at the same average velocity for twice as long it would go twice as far.

If it traveled at twice the average velocity for the same length of time it would go twice as far.

However it travels at twice the average velocity for twice as long, so it goes four times as far.

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Self-critique (if necessary): I had the first two parts of the problem correct, but not the distance travelled. I now understand that because the car was going twice the velocity for twice as long it will go 4x the distance.

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Self-critique rating #$&* 3

Question: `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length.

Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?

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Your solution:

The cord will stretch more than 7 feet.

confidence rating #$&* 1

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Given Solution:

‘aFrom 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft.

A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).

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Self-critique (if necessary): I got the answer right, but I did not have a good explanation. I graphed the points and it looked like 125 would be more than 7, but I did not know for sure. The explanation with the concave down nature of the graph helps, because the down curving graph puts the 125 point above 7.

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Self-critique rating #$&* 3

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Question: `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet.

The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force).

When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?

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Your solution:

She will go twice as far as the 4 ft push because the force is twice as great as before.

confidence rating #$&* 1

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Given Solution:

`aThe distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far

Self-critique (if necessary): I got that the distance was doubled, which would make the distance double, but I forgot that the force was doubled as well, with the bungee cord. This makes the distance double twice, so the skater goes 4 times as far.

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Self-critique rating #$&* 3

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Question: `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet.

To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first?

To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?

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Your solution:

The larger sphere will seem brighter because it is bigger, so there is larger surface area for the light to shine out of.

On the surface, the larger sphere will have less than half the brightness of the first. The light is more concentrated on the first sphere so the light in the 1 sq in will be brighter. The surface area of the second is 4x, so the concentration of light in the square will be smaller and it will be less than half as bright.

confidence rating #$&* 1

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Given Solution:

`aBoth bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area.

Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.

INSTRUCTOR RESPONSE: Imagine a light bulb inside a frosted glass lamp of typical size. Imagine it outside on a dark night. If you put your eye next to the glass, the light will be bright. Not as bright as if you put your eye right next to the bulb, but certainly bright. The power of the bulb is spread out over the lamp, but the lamp doesn't have that large an area so you detect quite a bit of light.

If you put the same bulb inside a stadium with a frosted glass dome over it, and put your eye next to the glass on a dark night, with just the bulb lit, you won't detect much illumination. The power of the bulb is distributed over a much greater area than that of the lamp, and you detect much less light.

INSTRUCTOR RESPONSE:

First you should address the explanation given in the problem:

'Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination. '

• Do you understand this explanation?

• If not, what do you understand about it and what don't you understand?

This simple image of a 2x2 square being covered by four 1x1 squares is the most basic reason the larger sphere has four time the area of the smaller.

There is, however, an alternative explanation in terms of formulas:

• The surface area of a sphere is 4 pi r^2.

• If r is doubled, r^2 increases by factor 2^2 = 4.

• So a sphere with double the radius has four time the area.

• If the same quantity is spread out over the larger sphere, it will be 1/4 as dense on the surface.

but less intensity at its surface, or the sphere with lesser area and greater intensity at its surface.

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Self-critique (if necessary): I got the second part, sort of, I figured out that the surface area of the bulb would make it seem dimmer than the smaller bulb, I just did not get that if would be ¼ as dense at the smaller. The area bit makes sense to me. The first part of the question, though, was difficult. I thought that you would be able to tell the difference, but now I know that the difference in area does not make a difference at a distance, only the energy from the light source.

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Self-critique rating #$&* 3

Question: `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius.

Place the following in order, from the one requiring the least energy to the one requiring the most:

Increasing the temperature of the ice by 20 degrees to reach its melting point.

Melting the ice at its melting point.

Increasing the temperature of the water by 20 degrees after all the ice melted.

At what temperature does it appear ice melts, and what is the evidence for your conclusion?

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Your solution:

1. Most - Melting the ice at its melting point.

2. Increasing the temperature of the water by 20 degrees after all the ice melted.

3. Increasing the temperature of the ice by 20 degrees to reach its melting point.

Ice melts at 0 decrees C because it already started to melt when the temperature reached 0 degrees and then the temperature could rise past 0 when all of the ice was gone.

confidence rating #$&* 2

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Given Solution:

`aSince the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius.

The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you.

Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fill in' the valley, with the result that you won't bob up and down very much.

If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point?

How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?

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Your solution:

When you are at the center the heights of the peaks will add to make it twice as high as normal (6in+6in) so you would bob up 12 inches and then go down 12 inches.

To cancel the waves out you need to displace a total of 3 feet. You have to move 1.5 feet from the center, which will give a total of 3 feet of movement. That change in position will put you in a place where the waves cancel each other.

confidence rating #$&* 3

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Given Solution:

`aIf the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in.

If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ]

However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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&#Good responses. Let me know if you have questions. &#

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