Greg has just started a new job at a grocery store down the road. ..

[eddy2017]

I get 12.92 mph you get 12.93 and you say it does math the 4 significant digit answer. I am not understanding what that means.

 

[Deleted member 4993]

I get 12.92 mph you get 12.93 and you say it does math the 4 significant digit answer. I am not understanding what that means.

Are you saying you do not know the "meaning" of the term "significant" digits?

 

[eddy2017]

I do know.
To determine the number of significant figures in a number use the following 3 rules:

1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
3. A final zero or trailing zeros in the decimal portion ONLY are significant

 

[eddy2017]

What I do not understand is what I did wrong in my dimensions analysis responding to the problem posted.

 

[Deleted member 4993]

The record time stands at 2:01:39 (hrs : mts : sec). The official distance is 42.195 kilometers.

Calculate Mr. Kipchoge's record making speed in mph (→ miles/hour ) with 4 significant digits.

1ft=0.3048m
1mi=5280 ft
1h =3,600 s

time = 2 hr 1 min 39 sec = 2.02750E+00 hr

Distance = 42.195 km = 42195 m = 42195/0.3048 ft = 1.38435E+05 ft = 1.38435E+05/5280 miles = 2.62188E+01 miles

speed = (Distance)/(time) = 2.62188E+01/2.02750E+00 = 1.29316E+01 mph = 12.93 mph

Now show your calculations...... and figure out where did you and I differed...

 

[JeffM]

Hi,
Greg has just started a new job at a grocery store down the road. He lives within walking distance of it. It takes him 20 minutes to walk there. He lives 2/3 of a mile from the store. Sometimes he rides his bike there instead, and it takes him only half as much time to get there.
Greg wants to know his speed in MPH when he is walking and his speed in MPH when he is riding his bike.
Calculate the speed in each case. Write your answers with walking speed first, followed by the bike riding speed.
WIK ( What I know)
it takes Greg 20 minutes to get to his job when he walks over.
it takes him half as much time when he bikes over=10 minutes.
speed in mph when walking?
speed in mph when riding his bike.

when he walks
S=d/t
S=(2/3)of a mile / 20 mts
S=92/3)/(20/1)
s=(2/3)(1/20)
S=(2)(1)/(3)(20)
S=2/60 reducing
S=1/30
S=0.033 mph
His speed when walking is 0.033 mph

when riding a bike,
S=d/t
s=(2/3)of a mile / 10 mts
s=(2/3)/(10/1)
=(2/3) (1/10)
=2/30
=1/5
=0.066
His speed when riding his back to work is 0.066 mph.


Is this ok?

Thanks in advance for corrections and tips.

The approach using dimensional analysis

You want to get an answer in miles per hour, which is distance over time. You have a distance and you have a time. WRITE IT DOWN.

[math]\dfrac{2/3 \text { miles}}{20 \text { minutes}} = \dfrac{2/3}{20/1} * \dfrac{\text {miles}}{\text{minute}} = \dfrac{2}{3} * \dfrac{1}{20} = \dfrac{1}{30} \dfrac{\text {miles}}{\text{minute}}.[/math]
But we do not want an answer in miles per minute. We want an answer in miles per hour. However, I know a relationship between hours and minutes. I want hours as a DENOMINATOR. So

[math]\dfrac{1}{30} \dfrac{\text {miles}}{\text{minute}} * 1 = \dfrac{1}{30} \dfrac{\text {miles}}{\text{minute}} * \dfrac{60 \text { minutes}}{\text{hour}} =\\ \dfrac{1}{30} * \dfrac{60}{1} * \dfrac{\text{miles} * \text{minutes}}{\text{minute} * \text{hour}} = 2 * \dfrac{\text{miles} * \text{minutes}}{\text{minute} * \text{hour}} .[/math]
Well that is a very weird set of units. But wait, I have minutes in numerator and denominator so I can cancel them.

[math]2\ \dfrac{\text{miles} * \cancel{\text{minutes}}}{\cancel{\text{minute}} * \text{hour}} = 2\text{ miles per hour.}[/math]
I am telling you that dimensional analysis prevents mistakes. Look at the video. There are plenty more videos out there. It is a very common technique because it is such a useful technique. It simplifies things by addressing the arithmetic and units separately, which prevents silly mistakes.

You should also check your work for reasonableness. 0.033 miles per hour! I have seen turtles that walk faster than that

 

[eddy2017]

This is very hard to follow.
I would do unit conversions before plugging values into formulas.
s = d/t.
d = 2/3 miles
t = 20 min = 1/3 hour.
Now plug them in:
(2/3)/(1/3) = (2/3)*3 = 2 mph

You missed an important step at the end - reality check. I understand that problems sometimes present very unrealistic scenarios, but if the givens are "normal", then the result should also be normal. In this case you got a bike speed of 0.066 mph from 2/3 miles in 10 min. Realistic givens, right? What about 0.066 mph? Is a bike that much slower than a car driving at 25-35 mph?

Yes, you're right.

 

[eddy2017]

The approach using dimensional analysis

You want to get an answer in miles per hour, which is distance over time. You have a distance and you have a time. WRITE IT DOWN.

[math]\dfrac{2/3 \text { miles}}{20 \text { minutes}} = \dfrac{2/3}{20/1} * \dfrac{\text {miles}}{\text{minute}} = \dfrac{2}{3} * \dfrac{1}{20} = \dfrac{1}{30} \dfrac{\text {miles}}{\text{minute}}.[/math]
But we do not want an answer in miles per minute. We want an answer in miles per hour. However, I know a relationship between hours and minutes. I want hours as a DENOMINATOR. So

[math]\dfrac{1}{30} \dfrac{\text {miles}}{\text{minute}} * 1 = \dfrac{1}{30} \dfrac{\text {miles}}{\text{minute}} * \dfrac{60 \text { minutes}}{\text{hour}} =\\ \dfrac{1}{30} * \dfrac{60}{1} * \dfrac{\text{miles} * \text{minutes}}{\text{minute} * \text{hour}} = 2 * \dfrac{\text{miles} * \text{minutes}}{\text{minute} * \text{hour}} .[/math]
Well that is a very weird set of units. But wait, I have minutes in numerator and denominator so I can cancel them.

[math]2\ \dfrac{\text{miles} * \cancel{\text{minutes}}}{\cancel{\text{minute}} * \text{hour}} = 2\text{ miles per hour.}[/math]
I am telling you that dimensional analysis prevents mistakes. Look at the video. There are plenty more videos out there. It is a very common technique because it is such a useful technique. It simplifies things by addressing the arithmetic and units separately, which prevents silly mistakes.

You should also check your work for reasonableness. 0.033 miles per hour! I have seen turtles that walk faster than that

Yes, you're right. I'll check it again. Thank you for the info!.

 

[eddy2017]

when he walks
S=d/t
S=(2/3)of a mile / 20 mts
S=(2/3)/(20/1)
s=(2/3)(1/20)
S=(2)(1)/(3)(20)
S=2/60 reducing
S=1/30
S=0.033 mph
Jeff, so the way I set it up is wrong there. Can you tell me if the mistake lies here?.
I know it is not a logical result so it is wrong .
So therefore it is wrong because this is the wrong approach, right?.

 

[eddy2017]

time = 2 hr 1 min 39 sec = 2.02750E+00 hr

Distance = 42.195 km = 42195 m = 42195/0.3048 ft = 1.38435E+05 ft = 1.38435E+05/5280 miles = 2.62188E+01 miles

speed = (Distance)/(time) = 2.62188E+01/2.02750E+00 = 1.29316E+01 mph = 12.93 mph

Now show your calculations...... and figure out where did you and I differed...

I didn't understand this, Dr Khan.
time = 2 hr 1 min 39 sec = 2.02750E+00 hr
Am I not supposed to take everything to seconds?
= 2.02750E+00 hr what is this?.

 

[Deleted member 4993]

I didn't understand this, Dr Khan.
time = 2 hr 1 min 39 sec = 2.02750E+00 hr
Am I not supposed to take everything to seconds?
= 2.02750E+00 hr what is this?.

Since the final answer is in hour - I chose to work with 'hour'. You may choose to work on seconds and finally convert to hour.

If you put your calculator in scientific mode - you will get numbers like:

XXXXX E 00- read in your calculator 's User Manual.

 

[eddy2017]

Since the final answer is in hour - I chose to work with 'hour'. You may choose to work on seconds and finally convert to hour.

If you put your calculator in scientific mode - you will get numbers like:

XXXXX E 00- read in your calculator 's User Manual.

Good. Lol. Thank you. Got it.

 

[JeffM]

S=d/t
S=(2/3)of a mile / 20 mts

[/QUOTE]
Perfect.

S=(2/3)/(20/1)

[/QUOTE]
What happened to the units? They disappeared. If you do not see them, how can you possibly keep track of them. The correct step using dimensional analysis is to separate the arithmetic and the analysis of units. The correct next steps are:

[math]\dfrac{2}{3} * \dfrac{1}{20} * \dfrac{\text {mile}}{\text {minute}} =\\ \dfrac{1}{30} * \dfrac{\text {mile}}{\text{minute}}.[/math]
But you do not want an answer in miles per minute. So what is the next step? Dimensional analysis involves multiplying fractions and never forgetting to write down the units in each step. It involves very little creative thought. I want to get rid of minutes and substitute hours in the denominator through cancelling. So do I want to multiply by (1 hour)/ (60 minutes) or by (60 minutes)/(1 hour)?

s=(2/3)(1/20)
S=(2)(1)/(3)(20)
S=2/60 reducing
S=1/30
S=0.033 mph
Jeff, so the way I set it up is wrong there. Can you tell me if the mistake lies here?.
I know it is not a logical result so it is wrong .
So therefore it is wrong because this is the wrong approach, right?.

 

[eddy2017]

Totally got it now.
By [math]60 minutes / hour[/math].

 

[JeffM]

Totally got it now.
By [math]60 minutes / hour[/math].

Yes!

 

[eddy2017]

Yes!

Gotta arrange the units I don't need in such a way they cancel out, leaving only the desired ones. The video explains really well and these vids are aplenty on the web. Just a question of practicing it and, like you say, never lose sight of the units.
I watched a video in which the tutor arranges every up with only the units first, and then fills out the empty spaces he leaves for the numbers. I like that because, from the start, you just have a set up of how your units are going to be and how you're gonna cancel them out.

 

[JeffM]

Gotta arrange the units I don't need in such a way they cancel out, leaving only the desired ones. The video explains really well and these vids are aplenty on the web. Just a question of practicing it and, like you say, never lose sight of the units.
I watched a video in which the tutor arranges every up with only the units first, and then fills out the empty spaces he leaves for the numbers. I like that because, from the start, you just have a set up of how your units are going to be and how you're gonna cancel them out.

There you go. Please show the url of the video you like. I’d like to watch it. Never too old to learn something new.

Now I am going to explain why the method works.

Suppose I want to turn a number of inches (let’s say a) into the equivalent number of feet. I name my variables for my GENERAL equation of equivalence:
[math]i = \text {number of inches, and } f = \text {number of feet}[/math].
My relevant equation of general equivalence in this case is
[math]12i \equiv f \implies \dfrac{f}{i} = 12 \text { and } \dfrac{i}{f} = \dfrac{1}{12} \text { assuming } f\ne 0 \ne i.[/math]Now I name my variables for my specific problem:
[math]a = \text {GIVEN distance when measured in inches, and } x = \text {distance measured in feet to FIND}.[/math]I set up my equation with units specified:
[math]ai = xf \implies x = \dfrac{ai}{f} = a * \dfrac{i}{f} = a * \dfrac{1}{12}.[/math]
The mechanics of dimensional analysis allow you a short cut through that algebra with little work and little chance of error. Learn to use it. It is a life-saver for this kind of problem.

 

[eddy2017]

There you go. Please show the url of the video you like. I’d like to watch it. Never too old to learn something new.

Now I am going to explain why the method works.

Suppose I want to turn a number of inches (let’s say a) into the equivalent number of feet. I name my variables for my GENERAL equation of equivalence:
[math]i = \text {number of inches, and } f = \text {number of feet}[/math].
My relevant equation of general equivalence in this case is
[math]12i \equiv f \implies \dfrac{f}{i} = 12 \text { and } \dfrac{i}{f} = \dfrac{1}{12} \text { assuming } f\ne 0 \ne i.[/math]Now I name my variables for my specific problem:
[math]a = \text {GIVEN distance when measured in inches, and } x = \text {distance measured in feet to FIND}.[/math]I set up my equation with units specified:
[math]ai = xf \implies x = \dfrac{ai}{f} = a * \dfrac{i}{f} = a * \dfrac{1}{12}.[/math]
The mechanics of dimensional analysis allow you a short cut through that algebra with little work and little chance of error. Learn to use it. It is a life-saver for this kind of problem.

Thank you so much. I'll study all that when I get home.

 

[eddy2017]

I looked for the video but did not find it. I might bump into it later. But I am pasting this one that i loved for future visitors of the thread or someone in neeed of a good dimension analysis explanation. i liked the way this tutor explained this.

https://
www.youtube.com/watch?v=DsTg1CeWchc