Sixten’s Perfect Math Class

To my cousin, who is in high school and asked me for help with this problem on his homework: Why are you asking me for help with math? I’m an art school graduate. Find the smartest person in your class (usually a quiet girl with glasses and long dark hair) and ask them to help you. Don’t laugh; it works.


Look for a girl like this. Such girls are often mathematical.

Anyway, I think this problem is simple enough so I can help you. Just keep my advice in mind for next time. It’s hard to deliver detailed explanations over the internet from all the way here in Seattle, especially when I’m trying to play Toho. Find someone in Chicago to help you.

(High school level)

Problem 25: A car is driven at a constant velocity of 25 m/s for 10.0 min. The car runs out of gas and the driver walks in the same direction at 1.5m m/s for 20.0 min to the nearest gas station. The driver takes 2.0 min to fill a gasoline can, then walks back to the car at 1.2 m/s and eventually drives home at 25 m/s in the direction opposite that of the original trip.

Draw a v-t graph using seconds as your time unit. Calculate the distance the driver walked to the gas station to find the time it took him to walk back to the car.

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The graph is broken up into four sections marked 1, 2, 3, and 4.

1: This is the person traveling at 25 meters per second (90 kilometers/hour, or a little over 56 miles/hour). The velocity is constant, so the graph of velocity is a horizontal line. This lasts for the first 10 minutes. The area under section 1 is a large rectangle 10 minutes (600 seconds) wide and 25 meters per second high. When you multiply the velocity by the time, you get the displacement.

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The distance covered during section 1 is 15 kilometers.

2: This is the person running out of gas and having to get out of his car and walk in the same direction, toward the nearest gas station. The problem gives him walking at a constant velocity of 1.5 meters per second (5.4 kilometers per hour, or a little under three and a half miles per hour.) He walks this way for 20 minutes. This generates another horizontal line, longer but much lower than the line of section 1. The area under section 2 is 20 minutes (1200 seconds) wide but only 1.5 meters/second high. As you would expect, because his speed is so much less without the car, he covers a lot less ground despite taking twice as much time as section 1.

The distance covered during section 2 is (1.5 * 1200) = 1800 meters, or 1.8 kilometers.

Two-minute break: There is a two-minute span in which the velocity is zero because the person is taking time to fill up his container with gas. Naturally, no distance is covered during this time.

3: Tired from walking and/or slowed by the weight of the gas, the person walks back to the car at a constant velocity of 1.2 meters per second (4.32 kilometers per hour, or a little over 2.5 miles per hour.) We know it’s a distance of 1.8 kilometers back to the car, but since he’s traveling at a slower rate, he won’t get there as quickly. He’s also walking back, so his velocity is negative. This generates a horizontal line that is lower than section 2, and therefore must be longer than section 2 in order to have the same area under it.

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To solve for the unknown time, we divide 1800 meters by 1.2 meters per second, arriving at 1500 seconds, or 25 minutes.

Compare the rectangles defined by sections 2 and 3. Section 2 is 1200 seconds wide and 1.5 meters per second high. Section 3 is 1500 seconds wide and 1.2 meters per second high. And both represent the covering of 1800 meters of distance.

4: In this section, the person realizes he has stuff to do at home and drives his car back the way he came at the same speed at which he got there. Section 4 is the mirror image of section 1, in that he covers 15,000 m (15 km) in 10 minutes by driving at the speed of 25 m/sec. However, he’s going in the opposite direction, so the line is in the negative position.

Draw a position-time graph for the situation using the areas under the velocity time graph.

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The graph is broken up into four sections marked A, B, C, and D.

A: The person travels at a constant speed of 25 meters per second for 600 seconds (10 minutes), covering 15,000 meters (15 kilometers).

B: The person travels at a constant speed of 1.5 meters per second for 1200 seconds (20 minutes), covering 1800 meters (1.8 kilometers).

Break: The person has a velocity of zero for two minutes, therefore his displacement remains constant (represented by a horizontal line).

C: The person travels at a constant speed of 1.2 meters per second for 1500 seconds (25 minutes), covering 1800 meters (1.8 kilometers) in the opposite direction.

D: The person travels at a constant speed of 25 meters per second for 600 seconds (10 minutes), covering 15,000 meters (15 kilometers) in the opposite direction.

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Note how the velocity and displacement graphs relate: The velocity graph is the slope of the displacement graph. The higher the velocity graph, the steeper the slope of the displacement graph. When the velocity graph is negative, the graph slopes downward. And when the velocity graph is zero, the displacement graph is horizontal, having a zero slope.


I hope this was helpful. You may ask me for help again in the future, but only if you can’t find someone in your class to help you. It doesn’t have to be a girl, although that is the optimal situation. When I was your age, I went to a boys-only high school and I still managed to find people to help with homework.

5 Responses to “Sixten’s Perfect Math Class”

  1. ExoByte says:

    Answer is
    Answer is
    0 seconds
    0 seconds
    ’cause there’s no gasoline in Gensokyo!

  2. CCY says:

    There are, in fact, other benefits to seeking out quiet girls with glasses and long dark hair as well, but I’m just biased.

  3. lolikitsune says:

    CCY, we don’t want to be conveying the wrong message to the delicate high school youths.

  4. Illusice says:

    Excellent work, Sixten! I’d give extra credit for a solution that beautiful! Ask him to introduce you to some of his friends’ older sisters.

    But the one problem is that you did the problem for him. Will he actually read through that solution carefully or will he mindlessly copy? Probably something in between. I used to tutor math in high school, and it’s very easy to get in a situation where the student is asking you what to write next, like a robot. Ideally, you’d want to establish what he knows, what he doesn’t, and help him mentally bridge that gap without doing the problem for him. In reality, compromises are made… lots of compromises…

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