topic 1.1 extended problem solving strategies

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Topic 1.1 Extended Problem solving strategies

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Topic 1.1 Extended Problem solving strategies. Topic 1.1 Extended A - Problem solving strategies.  One of the main “headaches” of physics has traditionally been the problem solving aspect of it.  Generally, a problem will be posed by someone else, and YOU will have to find the solution. - PowerPoint PPT Presentation

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Page 1: Topic 1.1 Extended Problem solving strategies

Topic 1.1 ExtendedProblem solving strategies

Page 2: Topic 1.1 Extended Problem solving strategies

One of the main “headaches” of physics has traditionally been the problem solving aspect of it.Generally, a problem will be posed by someone else, and YOU will have to find the solution.

Topic 1.1 ExtendedA - Problem solving strategies

Oftentimes at first glance, it will seem that there is not enough information. Or the problem may just seem impossible.A possible problem-solving strategy will be illustrated on the next slide. But the more practice you get, the easier problem solving will become. Don’t lose faith!

Page 3: Topic 1.1 Extended Problem solving strategies

Consider two trains approaching each other on the same track. When they are 10 km apart a fly, who can travel at 2 m/s leaves one train and flies to the other, continuing to go back and forth without resting until the trains collide. How far, in total, does the fly travel?

5 km/h5 km/h

10 km

Topic 1.1 Extended A - Problem solving strategies

FYI: A little thought may convince you that he reverses direction an infinite number of times. Thus he has traveled an infinite sum of progressively shrinking distances.

FYI: Our first “attack” might be a frontal assault, wherein we find a pattern for these distances, and sum them up. It can be done!

FYI: But an easier way is to use our d = rt formula:

FYI: A little thought may convince you that he reverses direction an infinite number of times. Thus he has traveled an infinite sum of progressively shrinking distances.

Page 4: Topic 1.1 Extended Problem solving strategies

Consider two trains approaching each other on the same track. When they are 10 km apart a fly, who can travel at 2 m/s leaves one train and flies to the other, continuing to go back and forth without resting until the trains collide. How far, in total, does the fly travel?

5 km/h5 km/h

10 km

Topic 1.1 Extended A - Problem solving strategies

Since the trains are each moving at 5 km/h, they approach each other at 10 km/h.Since they are 10 km apart when the fly begins, he has exactly 1 hour to fly.That is to say, he has 60 minutes, or 3600 seconds to fly.Since he is flying at 2 m/s, we can use the formula d = rt to get the total distance of 7200 m, or 7.2 km.

Page 5: Topic 1.1 Extended Problem solving strategies

Topic 1.1 Extended A - Problem solving strategies

Here is a possible problem solving strategy:

List and label

Draw a picture

Determine attack angle

Determine relevant equations

Manipulate equations

Do unit conversions

Do calculations

Check reasonableness

Do the PHYSICS

Do the TRANSLATION

Do the MATH

Do the CHECK

The trains, the arrows and the labels all fit in this category.

5 km/h

Use d = rt.

Substitute and solve.

No manipulation of equation is needed, since we are looking for d.

Does the answer seem reasonable?

Don’t try infinite sum.

FYI: We used unit conversions after our translation, demonstrating that the particular order shown is NOT the only order. BE FLEXIBLE, and develop your own strategy.

FYI: Sometimes a check can’t be done – after all, what is reasonable here? But usually, an estimate can convince you that your answer is in the ball park.

Page 6: Topic 1.1 Extended Problem solving strategies

Topic 1.1 Extended A - Problem solving strategies

What is the area of a circle in m2 if it has a diameter of 1.25 cm?

Draw a picture Determine attack angle

Determine relevant equations

Manipulate equations

Do unit conversions

Do calculations

Check reasonableness

D = 1.25 cm

D = 1.25 cm · 1 m100 cm

D = 0.0125 m

List and label

Find R then use the area formula.

D = 2RA = πR2

R = D ÷ 2

R = D ÷ 2R = 0.0125 ÷ 2

R = 0.00625 m

A = πR2

A = π·0.006252

A = 0.000122718

A = 0.000123 m2

Actual size