2 trains going E/W and 65mph/85mph pass each other, how long before they are 330mi apart?

How to solve an algebra motion word problem.
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[@toughenupfluffy7294]

Two hours and twelve minutes. That's how long before you realize you're on the wrong train.

[@devonwilson5776]

Greetings. The answer is 2.2 hours. We will determine the solution to this problem by using the formula D=VT in which D is the distance to be covered, V the velocity, rate of change of distance with respect to time, and T the time taken to cover the distance to be covered, travelled. Now, we must first realize that the rate of separation of the trains will be the individual speed of each train added together, (85+65) MPH. That is, the trains are separating from each other at a break neck speed of 150 MPH. Now, we are to determine how long this rate of separation will take for the trains to be 330 miles apart. The time will be determined by using our D=VT given above, and plugging in the known values gives 330=150T from which it is determined that T=330/150, and T=2.2 hours.

[@kennethwright870]

Rate*Time=Distance Combined Rate is 65mph+85mph=150mph, so 330mi/150mi/h = 2.2h

[@Snap-Anzahl]

Wouldnt it be faster to add the speeds together and divide them by 60. This would have them traveling at 2.5 miles per minute then divide the distance (330) by 2.5 = 132 min.

[@gennyshovel7083]

2hr 12 min

[@chuckmoore8219]

2 hours and 12 minutes

[@jajapani]

Love your videos, but for god sake, can you do it quick.

[@Tokinjester]

The whole point of the video is to teach people who don't know math, they need the explanation...just click on the end of the video to see if you're answer's correct, that's what most people do 🤷🏻‍♂️

[@chuckmoore8219]

Too long winded

[@dlkline27]

Got 2.2 hours working in my head. Took about 15 seconds.

[@deech18]

Right? People here proud of doing things the long way

[@williamniver6063]

I did, too. I would infer from both of our experiences that we are of a similar age. In the '60s and '70s I attended a very traditional boys' prep school. Hours of drills, loads of homework EVERY night, including weekends, and an embrace of rote memorization. Sounds barbaric, doesn't it? Requiring 12 and 13 year-olds to study Latin before taking on a modern language. Algebra in 7th grade, allowing for two years of calculus before going to college. No real "electives" either, no music , art, or drama classes EVER in eight years (5th - 12th), but you could choose a second year of biology or organic chemistry or physics on senior year, and English "electives" might be a choice between "17th Century French Dramatists' or "The Lost Generation: Fitzgerald, Hemingway, and Stein". But psychology, technology, anthropology, fuggetaboutit. Discipline was swift and effective: talk out of turn in class? 2" masking tape over my mouth at least for that class, but a second time = taped til lunch. Heard to curse or being sneaky breaking rules or wilfully disregarding standards? One of three paddles smacked across your ass, whether you were 10 or 19. Have a beef with another student? The combatants (and most the rest of their class) taken to the gym, strapped on (huge) boxing gloves, and the other kids forming a "ring" with one of the coaches as referee. Issues were resolved, then declared over, no further repercussions, combatants required to shake hands and publicly announce resolution of hostilities. On the other hand, there was the Junior-Senior smoking room, anyone could buy fresh jolt coffee for a dime before class, nearly every 10th - 12th grader free to leave campus at lunch, and minor things like underage drinking or seducing girls from the local public high school - and a few times the female faculty - went ignored, do long as you kept up your grades. The result: 100% of graduates accepted at four-year colleges or universities. Every year. For decades. Not one exception. And for the last 59+ years, I have never heard a single complaint from an alum. Not a peep. We DO decry the current day additions of art and music classes at the expense of woodshop, the explicit rejection of corporal punishment, and namby-pamby tobacco-free campuses. Tuition in 1970 was $1,650; in 2023, over $46,000. Did I mention doing most of these math review problems in my head?

[@aebalc]

I would give that answer a fail. No one tells time using a decimal system. As the problem is worded you would need to answer 2 hours 12 minutes.

[@Tokinjester]

​​@@aebalc there is a business discipline that uses the decimal system to calculate time tho' I used to be a Work Study officer and our stopwatches are marked with 100 _seconds_ per minute ...it makes calculations easier Also, 2.2 minutes is exactly how I calculated it 😉

[@jacoblussier999]

What do you want a fucking cookie

[@danieldennis9831]

It will be 2.2 hours or 2 hours, 12 minutes as long as the trains are in the middle of the US because only there are the train routes that straight and flat for such long distances. (65+85)x=330 150x=330 x=330/150 x=2.2

[@panlomito]

That was my idea too: time = distance / speed (in units: hours = miles / miles per hour = miles . hours per miles = hours ) So time = 330 miles / (65+85 mph) = 330/150 = 33/15 = 11/5 = 2 1/5 = 2.2 hours = 2 hours plus 60 minutes . 2/10 = 2 hours and 12 minutes Units are important !

[@terry_willis]

It depends how long the trains are. Some freight trains are 1 or 1-1/2 miles long. When you ask "how far apart" which part of the trains are you referring to? The front of the engine or the rear of the caboose (or last car)? Or somewhere in between?

[@richardowens9061]

They have not actually passed each other until the trains are no longer passing each other - which means, the rear of the last car of each train has to pass each other. How far apart the engines are is irrelevant for this particular word problem. The distance between the trains would necessarily be end to end. Of course, the word problem could have been made more complex by specifying the length of each car, the length of the engines, and how many cars in each train, the standard distance between each car, and which train was going what speed, and whether the distance should be between the engines or between the end of the last car of each train. It could further be complicated if the two trains began at rest beside each other and accelerated at different rates in each direction up to their stated speed. You could even introduce varying slopes for each train that reduced or increased their speed. The potential variables are, essentially, endless.

[@joekauffman2884]

2 hours and 12 minutes between Texarkana and Little Rock.

[@m.h.6470]

Solution: Everything is relative... so it doesn't matter, if you have two moving trains with 65 and 85 mph or one moving train with 150mph (=65 + 85) and one stationary train. So we simply need to calculate, how long it would take a 150mph train to reach 330 mile: 330 mi/150 mi/h = 330/150 h = 2.2 h = 2 hours and 12 minutes

[@nicholasb8900]

Your formula is slightly off. It should read 330 miles / (150 miles / hour) which equals (330 miles hour) / (150 miles) to get the proper units

[@m.h.6470]

@@nicholasb8900 mi/h is the same as mph. it is one term. I only wrote it that way, so that it becomes clear, that mi cancel out and hours remain.

[@nicholasb8900]

@@m.h.6470 yea I was being too nit picky. I blame it on the math KZbin channel I been binging on who harps on units.

[@MFisher7346]

Hours are not metric. 0.2 hours = 12 minutes.

[@luillierstephane1463]

2.2 h ??!! I certainly wouldn't write it this way. I found 02:12 , 2 hrs and 12 min.

[@GoofballFlyer]

I use tenths of hours when recording my flying hours. The clock (aka Hobbs meter) in the airplane records time that way to make adding time easier than hh:mm. 0.2 hours = 0.2 x 60 = 12 minutes.

[@Michael43713]

Got in 3 mins. Freeze one train then relative speed = 150mph. Then time = distance/speed.

[@thomasharding1838]

Trains usually go north-east (or north-west) (n/e or n/w) or south-east (or south-west) (s/e or s/w) but usually east/west would be called North (or South). So, because the trains are traveling at a differential speed of 20 mph, it would take them 16.5 hours to open a distance of 330 miles between them. UNLESS, they are going in opposite directions! Then it would take 2.2 hours, because their differential speeds would be 150 mph. But their relative direction (same or opposite) is not stated, is it‽ (The preceding assumes common fixed points of measurement on each train, Engine to Engine, or Caboose to Caboose, and not the air gap between the leading Caboose and trailing Engine.)

[@GailUnderhill]

Since you’re a math teacher, who uses the word “problem” so often; perhaps you could pronounce it correctly. It would be easier to listen to you speak. In your defense, you are probably tired of saying the word “problem”. However, we must listen to it repeatedly. Occasionally, you do pronounce it correctly. I admire your math skills. I struggle with algebra. However, I do enjoy grammar, literature, reading, foreign languages, history, geography, and spelling. God gave each of us talents, interests, and aptitudes for learning. Math is not an easy subject for me. I enjoy how you explain, in detail, the procedures to solve the problems. Thank you for your help! I really like your channel. With so many subscribers, it is clear that many people need help with math.

[@Real_LiamOBryan]

I don't know why, but my brain thought .2 of an hour, obviously 1/5th of an hour, was 6 minutes, obviously 1/10th of an hour. This led me to think the answer was 2 hours and 6 minutes. Fail, brain.

[@kennethstevenson976]

A more useable answer could be found by converting 330/150 = 33/15 = 2 12/60 = Two Hours and 12 minutes.

[@GoofballFlyer]

This assumes the trains maintain their speed at the time of passing. That's highly unlikely, so state the assumption.

[@GillianWainwright]

Another case of this guy being unclear. If you got this on an exam as it is written, you wouldn't know if they were both going east to west or in opposite directions.

[@HeatherDiMarco]

2hr and 12 min assuming they are both really short trains and pass each other very quickly

[@ernestannang9435]

What if they are moving towards each other? What Will be speed(rate)?

[@DominicMazoch]

Suppose onr or both trains reach their endpoints before 385 miles?

[@renenadeau323]

One train going 150mph. 330/150= 2.2hrs.or 2hrs 12 min.

[@raynewport9395]

We are only given the passing speed, so we have not enough information.

[@ec633]

Fourteen minutes

[@Real_LiamOBryan]

Yep.

[@loneranger6361]

330/(85+65)=2.2 hours

[@keithschipiour4684]

This why I was never good at math . Because I dont care about any stupid trains or how far there apart . Same reason I am not a expert on monkey sex life .

[@Tokinjester]

Same reason you're not an expert in English too? [ *_they're_* apart ] 😂

[@DelmarBrewer]

2hours and 12 minutes

[@Mrskateboardboy]

2 hours and 12 minutes

[@williamkirkconnell7306]

16

[@andypandy9931]

2 hours 12 minutes

[@tommykyle1570]

Easy problem

[@kennethhubbard2027]

2 hours 12 min

[@johnallen3555]

how long are the trains? - how is the apart mesuresd? (along the rail line or in a straight line) - if in a straight line, the curvature of the earth should be taken into consideration - We would need a map of the rail line as the rail line is most likely got curves etc in it. - Now, only the passing speed of the trains is provided, I am not aware of any piece of straight track with no stations that will allow for full unadjusted speed for the required distance. So, we likely do not know the average train speeds over the 3 hours By ripping in without thinking, numnuts comes up with an answer that, were the same process be applied in engineering would likely result in some one being killed or some other major stuff up.

[@JonBernhards66]

Dude, wake up! The earth has no curvature to measure! Earth is a topographical plane.

[@richardowens9061]

Both trains are on tracks. So, regardless of whatever curves might be involved, the distance is necessarily going to be measured on the tracks. And, since the speeds are specified and remain constant, acceleration can be dismissed. And, the distance between the two trains is specified as 330 miles. Remember, we have only the information specified in the word problem - so, that is all we have to consider to come up with a correct answer.

[@andrewwhite1038]

2hrs 12mins

[@keithschipiour4684]

Somebody forgot throw the switch trains were on same track had head on collision and did not pass each other .

[@richardfrankham7522]

Probm 😅

[@richardkanavel4358]

2,2 hrs

[@MathsMadeSimple101]

They probably had to stop at multiple stations each. So impossible to answer

[@russelllomando8460]

got it. in minutes 2 hr 12 min 12 min is .2 hours - 1/5th.

[@agentD99]

2 hours 15 minutes give or take.

[@mylittlepitbull3143]

Real quick in my head I got 330/150 hours Which really is not too bad if you think about it. I might take the train next time.

[@freeguy77]

In Oct-Nov. 1966, I took the Last Train to Clarksville!

[@mylittlepitbull3143]

@@freeguy77 did it take 2.2 hours

[@Elvis.818]

2 Hours 12 min

[@JoeGator23]

You asked the question wrong... I thought they were both traveling in the same direction, not diverging.

[@StephenRayWesley]

65mphx+85mphx+330miliesx+60miliesx=65mphx-85mphx-330miliesx-60miliesx= 130mphx+170mphx660miliesx +120minutresx = {1080mphxmphxmiliesxminutesx÷9}= 120 (x+9x-120)

[@adamglltt89]

Add the speeds together. Divide the miles by the speed. 2hrs and 2 10ths of an hour. 2h12 minutes. Speed Distance Time triangle helps with these things.