Finding average speed (hopefully this belongs here)
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My calc prof. sprung this one on us for fun one day, and most of the class fell into the expected pitfall. I heard even Einstein fell for it. I've used variations on it for years to watch most of my friends fall for it too.
Steven G
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I knew not to fall for just taking the average routine but did not see what was going on for a while. Actually I spent most of my time rechecking my work but after tkhunny hinted i was right then I immediately saw what was going on--just differently then you saw it.
Otis
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If you were to fix your incorrect value for t1 and finish solving for t2, you would get t2=0.
That result tells us the given scenario is impossible because no object can move 0.25 miles in 0 hours.
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Steven G
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I have learned over the years to never accept the obvious and look closely at all problems. I have been burnt too many times and try by very best not to let that happen any more. Of course it still does just not as often!My calc prof. sprung this one on us for fun one day, and most of the class fell into the expected pitfall. I heard even Einstein fell for it. I've used variations on it for years to watch most of my friends fall for it too.
Otis
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A138, it's what we call a "trick question" because the question is worded as though a numerical solution exists, when there is none.
Here's a simpler situation. You go 5 miles to the store and then return. It took one hour to reach the store, so your rate was 5 mph (going to the store). How fast does the return trip need to be, to average 10 mph for the round trip?
10 mph means moving 10 miles in one hour. Well, the round trip is 10 miles, but you used the entire hour just getting to the store, so there is no time remaining for the return trip. In other words, it's impossible to average 10 mph, in the given situation.
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That's my thinking and technically "common sense" tells me the answer is correct (ABSOLUTELY incorrect .... read the responses above) ....so yes your correct....but apparently we didn't make it complicated enough for everyone else....ugh.
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Otis
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Hi A138. I woke up with another thought, about your note that we seem to be speaking a foreign language in discussing the math. I'm also curious about the assignment, in which you said there was only one distance/rate/time question (and it was optional). How long has it been since you worked on such word problems?
If you're not very familiar with the relationship d=r·t (and its other versions), then I'd like to know more about your assignment. I'm starting to wonder whether it was some type of placement test.
If you're interested, we can help you understand Jomo's approach (setting up and solving the equation I'd posted in red). Let us know.
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If you're not very familiar with the relationship d=r·t (and its other versions), then I'd like to know more about your assignment. I'm starting to wonder whether it was some type of placement test.
If you're interested, we can help you understand Jomo's approach (setting up and solving the equation I'd posted in red). Let us know.
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Not a placement test, just an extra credit question. My professor is a funny guy, none of the other questions had anything to do with distance calculations. I read your store scenario and it made sense that the answer is 0 because there wasn't enough time. But, yes! If you want to help me understand his approach, I'd like that.
Otis
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Don't feel badly, creelrj. As I mentioned earlier, Einstein made the same goof.
Also, a correct response to the exercise is not complicated to formulate. I extend the same invitation to you, as I did to A138. We can go over the exercise again, in more detail. Let us know.
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Otis
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Yes, you mentioned that. I'm interested in knowing whether the class studied any distance/rate/time exercises. If not, how long has it been since you did? I can go over the setup and solving, but I'd like to know more of what you've already learned about d=r·t problems and their other versions.
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Otis
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Hi A138. I can't know for sure what your funny professor had in mind. I'd like to explain how Jomo and I approached the exercise. But first, let's cover some basics.
When an object moves, there are three related quantities: distance traveled, speed (also called 'rate') and time duration. Distance may be expressed as the product of the other two quantities. We can write the relationship as:
distance = rate × time
Using symbols d, r and t to represent these amounts, we write:
d = r · t
This is a formula for calculating distance, when we know both rate and time.
There are two other versions of this basic relationship. We can solve that equation for t (divide each side by r). That gives a formula for calculating time, when we know both distance and rate:
t = d / r
We can also solve d = r · t for r (divide each side by t). That gives a formula for calculating rate, when we know both distance and time:
r = d / t
Many beginning exercises provide two of the three quantities, and the student calculates the missing quantity (by using the appropriate formula above).
For example, if we're told that an object moves 23 miles in half an hour, then they've given us the distance and the time; it's the rate that's unknown. We use the formula for rate (dividing distance by time). That generates the unit 'miles/hour' because miles is the numerator and hour is the denominator.
r = d/t = 23/0.5 = 46 miles per hour
If we were given the distance and rate, instead, then the time is unknown. We use the formula for time (dividing distance by rate), and the resulting unit is hour.
t = d/r = 23/46 = 0.5 hour
Here's what the unit cancellation looks like, when dividing a length (miles) by a rate (miles/hour). Remember from arithmetic that dividing by a ratio is the same as multiplying by its reciprocal (in other words, we don't do division with fractions; we switch to multiplication, using their reciprocal).
\[time = \frac{distance}{rate}\]
\[\quad \quad \; = \;\; \frac{miles}{\frac{miles}{hour}} \;\; = \frac{\bcancel{miles}}{1}× \frac{hour}{\bcancel{miles}} = \frac{hour}{1}\]
See the reciprocal? Instead of dividing by miles/hour, we multiply by hour/miles.
Students are expected to understand and memorize the formula d = r · t (the other two versions can then be derived, by solving the equation for r and for t).
Students also need to be consistent with units. If an exercise gives distance as kilometers and time as seconds, yet asks for a rate in miles per hour, then the distance unit is 'miles' and the time unit is 'hour', so the student needs to first convert kilometers to miles and seconds to hours.
When an object changes speed along the way, then we have different rates for different segments of the entire trip. Therefore, each segment has its own distance and time, too. We may then calculate an average rate for the entire trip. The formula for average rate is similar to r=d/t, but we use the total trip distance and the total trip time:
average rate = (total distance) / (total time)
In this thread, the exercise has two segments: running the first lap and running the second lap. Symbolically, we can use subscripted variables to distinguish the separate quantities for each segment. Again, each segment has its own distance, rate and time:
Let d1 = distance in lap 1
Let r1 = rate in lap 1
Let t1 = time in lap 1
Let d2 = distance in lap 2
Let r2 = rate in lap 2
Let t2 = time in lap 2
Using these symbols, we express a formula for the average rate (total distance divided by total time):
average rate = (d1 + d2) / (t1 + t2)
Let's think about what we know. We're given the distance of a single lap: 400 meters. We need to convert that to miles because the exercise uses 'miles per hour' for rates. 400 meters is roughly one-quarter of a mile. Therefore:
d1 = 0.25
d2 = 0.25
The exercise gives us r1, also. It's 5 miles per hour. That's enough information to find t1. We use the time formula and substitute the known values for the distance and the rate:
t1 = d1 / r1
t1 = 0.25 / 5 = 0.05 hours
Now, the exercise asks us to find the value of r2 that would make the average rate (for a two-lap trip) 10 miles per hour. How can we express the rate r2? We use the rate formula:
r2 = d2 / t2
We know d2 = 0.25, so we make that substitution:
r2 = 0.25 / t2
At this point, we realize that we need the second time t2, to calculate the requested rate. How do we find t2? By writing and solving an equation, and that equation is for the average rate because it also contains symbol t2. We know all of the other values, to be able to solve for t2
average rate = (total distance) / (total time)
average rate = (d1 + d2) / (t1 + t2)
We substitute all the known values:
10 = (0.25 + 0.25) / (0.05 + t2)
The first step to solve that equation is doing arithmetic that's possible. We can add 0.25+0.25:
10 = 0.5 / (0.05 + t2)
The next step is to get t2 out of the denominator. Multiply each side by the denominator:
10(0.05 + t2) = 0.5
The next step is to get rid of the grouping symbols. We use the distributive property, to carry out the multiplication on the left-hand side:
0.5 + 10 · t2 = 0.5
With enough practice solving equations, we see by inspection that t2 must be zero. But, if we don't see it, then we finish solving by subtracting 0.5 from each side and then dividing each side by 10 (to get t2 by itself on the left-hand side):
10 · t2 = 0
t2 = 0/10 = 0
That result is a contradiction because it says running the second lap requires zero time (which is impossible). Also, look what happens, if we use that result to report the requested rate r2:
r2 = d2 / t2
r2 = 0.25/0
The number 0.25/0 does not exist because division by zero is undefined.
These contradictions tell us that an average rate of 10 mph is not possible.
Questions?
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