Average Velocity Problems - Easy
Potential Pitfall!:
Remember that there is a difference between average velocity and average speed. Average velocity is the change in position (final position minus initial position) divided by change in time and is a vector. Average speed is the total distance traveled divided by change it time and is a scalar.
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Remember the Basic Principles
1. Draw a picture. 2. Write down what information you have and what you are looking for. 3. Figure out what information is implied 4. Determine what formula(s) apply. 5. Solve the equation for the missing information 6. Check your answer |
John Walks to the Park Part 1: John wants to go the the park that is 2 kilometers from his house. He walks at 3 mi/hr. How long will it take him to get to the park walking at that velocity? Step 1 is "Draw a picture." Of what? Of John walking to the park. Here is an example rendition of John walking to the park. Please note the stick figure and unrealistic scale. Also note that it doesn't matter! Assign a coordinate system and choose a frame of reference. In this case, we choose the problem to start at the point when John leaves his house. And note that this is a one-dimensional problem - we are only concerned about John's movement in one direction, in this case the x-direction.
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Potential Pitfall! - Funny Units The units of distance are kilometers while the units of velocity are in miles per hour. To avoid getting the wrong answer, always covert to standard SI units before doing the calculation. |
Step 2 is writing down the information we have:
..and the information we want to find:
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| Step 3: What information is implied? In this case, the implied information involves converting from non-standard units to standard SI units. So, we have to covert from kilometers to meters and from miles/hour to meters/second. | |||||
| For help with conversions, click here |
kilometers to meters --> 2 km x (1000 m)/1 km = 2000 m. miles/hour to meters/second --> (3 mi/hr) x (1hr / 60 min) x (1min / 60 s) x (5280 ft/1 mi) x (12 in/1ft) x (2.54cm/1in) x (1m/100cm) = 1.34 m/s |
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Step 4: What formulas can be used to solve this problem? Review the six
kinematic equations. The simplest one to use given the information we have
is
v = s/t We have velocity (1.34 m/s), and we have distance (2000 m), so now all we have to do is solve for time! |
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| Step 5: Solve the equation. Solving for t
gives:
v = s/t tv = s t = s/v Now, plugging in our values of v and s: t = (2000 m)/(1.34 m/s) Collecting values and units: t = (2000/1.34)(m/m/s) Meters cancels, leaving our answer in units of seconds, which is what we would expect when solving for time. t = 1.493 x 103 s This answer means very little to us because we are used to dealing in units of minutes or hours. However when answering homework or test problems you should always leave your answers in SI Units unless otherwise specified. Also, always use scientific notation. But for the sake of this problem, we will convert back to minutes and see if the answer makes sense. |
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| Step 6: Does it make sense? Converting to
minutes, we see that it take John roughly 25 minutes to walk the 2 km to
the park.
(1.493 x 103 s) x ( 1 min / 60 s) = 24.87 min. 2 km is approximately 1.25 miles. At a rate of 3mi/hr, John should cover one mile every 20 minutes. Making a rough calculation, John should travel the distance to the park in 25 minutes. Thus the answer for time that we calculated is reasonable. |
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| Part 2: After John is at the park for half an hour, he discovers that he dropped his sunglasses. He retraces his steps and finds them .80 km from his house. Assuming that he walks at the same speed to find his sunglasses as he did while walking to the park, what is John's average speed? | |||||
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Step 1: Modify the picture to fit the new problem. John travels 2 km
to the park and travels back from the park to .80 km from his house.
Please note that this image shows the entirety of the problem - not just a single point in time. It shows John walking to the park (x= 2.0 km) and it also shows him traveling to the point where he dropped his sunglasses (x = .80 km). |
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Steps 2 & 3: The information we have now consists of all relevant
information from Part 1. The information we want to find is now the
vector quantity of average velocity. Remember that velocity is change
in displacement divided be change in time. So we need to find two
different quantities: 1) What is the total displacement from John's
house? and 2) how long did it take him to get there? Once we have
these values we can solve for average velocity.
So now we know from Part 1 that:
And we are told in Part 2 that:
What we need to find is John's Average Speed. We need to know the total displacement or change in space (s) and the total time (t) or change in time it took him to get to that displacement. Which means this problem now has two values for s and two values for t. To avoid confusion, we will label all values from part one with a subscript of 1 and values from part 2 with a subscript of 2. |
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Step 4: The formula for average speed is v = s/t In the case of average speed, which is not a vector but still has the symbol v, average speed is total distance traveled divided by total time.
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Step 5: Plug it in. First we need the time it took him to find his
sunglasses. The same formula that lets us solve for average velocity
will let us solve for the time. When John walks back to find his
sunglasses, he travels an additional distance of 1.2 km or 1200 m. (He
leaves the park at x = 2000 m and ends at x = 800 m. The difference
between these is 1200 m.) So after leaving his house, the total
distance he has traveled is
2000 m + 1200 m = 3200 m. Because John walks at 1.34 m/s and travels 1200 m back to his sunglasses the time it take him to get to his sunglasses is v = s/t Solving for t gives t = s/v Plugging in and solving as in Part 1 t = (1200 m) / (1.34 m/s) t = 896 s To find the total time, add together all values for time 1493 s (to the park) + 1800 s (at the park) + 896 s (looking for sunglasses) = 4189 s or 4.189 x 103 s Since Average Speed is total distance, which we found to be 3200 m, divided by total time, which we found to be 4.189 x 103 s v = s / t v = (3200 m)/(4.189 x 103 s) v = .763 m/s So John's average speed is v = .763 m/s. Step 6: Is this reasonable? Well, his average speed should be less than his walking speed, because he spent time at the park without traveling any distance and this time is calculated into the average speed formula. So yes, this is a reasonable answer.
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| Part 3: After finding his sunglasses John
decides to go home. When he arrives at home, what was his average velocity
for the entire trip?
Following our procedure as outlined twice above, we end up with a picture that looks like:
The Formula we want to use is v = Δ s / Δ t Or, the change in position divided by the change in time. But remember that John travels 2000 m to the park and 2000 m back home. His initial position is at x = 0 and his final position is at x = 0. Δ s = sf-si Plugging in, we end up with Δ s = 0 m Now plug into the kinematic equation for velocity v = Δ s / Δ t v = 0 m/s Note that is doesn't matter how long it take John to get home; because his total displacement was 0, so too is his average velocity. And that, ladies and gentlemen, illustrates the difference between average speed and average velocity |
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Now You Try It 1. How much time did John spend away from home? 2. Johns average velocity was calculated to be zero upon his arrival home. But what was his average speed for the entire trip?
© 2006, Spitfire Science
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