When will Car B catch Car A after a delayed start?

Geoff Wales · 2011-02-08 17:11:55

Car A leaves at 35kph. After 45 minutes Car B leaves at 50kph. When will Car B catch Car A?

What I need is the solution, rather than the answer.

Thanks

Geoff

bobbym · 2011-02-08 17:20:12

Hi Geoff Wales;

Geoff Wales · 2011-02-08 17:28:18

I'm not a maths (math) teacher, although I do teach some basic maths in Primary (Elementary) School. I'll see if I can follow your explanation, but as yet I don't understand it.

I will need to come up with a simpler solution if I am going to show Year 6 how to solve this.

Thanks for the quick reply. What a great place to find!

Last edited by Geoff Wales (2011-02-08 17:31:24)

bobbym · 2011-02-08 17:30:36

Hi Geoff Wales;

What part of the explanation did you have problems with?

Geoff Wales · 2011-02-08 17:36:19

Can you explain in words what you have written in math terms?

What is t in 35t? Time?

I know S = D/T, so D = SxT, right?

Last edited by Geoff Wales (2011-02-08 17:36:46)

bobbym · 2011-02-08 17:39:54

That is right! Do not let capital letters confuse you. d is distance in your formula. distance = rate x time

Geoff Wales · 2011-02-08 17:46:02

What is the dot in 35.3/4? 35 times 3/4?

bobbym · 2011-02-08 17:48:53

Yes, that is a multiplication. 45 minutes is 3 / 4 of an hour. We know A's head start of 45 minutes means it will take B some time to catch him. But common sense tells us that he will because he is going faster.

Another way to look at this is that A is going 15 kph faster than B. B has a 45 minute head start at 35 kph. How long will it take A to cover that 35 * ( 3 / 4 ) k head start. So you could solve it like this:

Bob · 2011-02-08 21:37:05

hi Geoff Wales

I've only had a quick glance at this but it looks like a suitable case for a distance time graph. That should be easier for your class.

I'll have a longer look later when I get back and post a graph if you need it.

later edit.

It is. see below.

Bob

Last edited by Bob (2011-02-08 22:08:41)

Geoff Wales · 2011-02-09 00:01:29

Yes, I did draw a graph to help me visualise the problem.

Can I assume that the 35t assumes a time of one hour?

So, '35/60 x t' becomes 35t, right?

This is not an easy question for this age group. I want to make it as simple as possible.

Car A travels at 35 kph plus a head start of 35kph for 3/4h

I need to explain how I get from s = d/t to d = 35t, so 35km/1hour helps to simplify the term to 35, not 35/60.

35kph = d/t

d = 35kph x t (simplify by ignoring 'kph')

35t plus the extra 3/4hr at 35kph = 50t

(We don't write 35 x 3/4t, because 't' refers to an unknown time, where the 3/4hr is a known time - is that right?)

15t = 35 x 3/4

t = 35/15 x 3/4

t = 7/3 x 3/4

t = 7/4

7/4 hours from zero hours, or 10/4 hours from zero hours?

Is it 1 3/4 hours from the start of the race, or 2 1/2 hours?

I see you asked the question, "how long will it take Car B to catch up with Car A" (you got the A and B mixed up, I think), so that would imply we have to add the 3/4 hour to find the time from the start.

Last edited by Geoff Wales (2011-02-09 00:08:19)

Bob · 2011-02-09 00:07:29

hi Geoff

I see you are on line. Stay on if you can and I'll talk you through it

bobbym started his 'time' when the second car leaves. I started mine from the first car, so our answers will be 45 mins apart.

His is 1.75 hours. My graph shows 2.30 hrs which is 1.75 plus another .75.

(note 45 mins is .75 hrs)

Bob

Last edited by Bob (2011-02-09 00:10:53)

Geoff Wales · 2011-02-09 00:11:23

I am getting it, slowly. If you can answer a couple of the question I put, that would be great.

My main problem is understanding the way the terms are used, I think. I have trouble seeing the connection between the symbols (and the way they are used), and the problem itself.

EDIT

Yes, that was my understanding, too.

Last edited by Geoff Wales (2011-02-09 00:12:29)

Bob · 2011-02-09 00:12:18

hi

Are you teaching in the UK?

That's why I think ythe graph will help. I always try to draw pictures to help with tricky maths.

Bob

Last edited by Bob (2011-02-09 00:15:19)

Geoff Wales · 2011-02-09 00:14:13

I'm in Sydney, teaching selective school trial test Maths.

They do practice papers for one year before the actual test, in Maths, English and General Ability.

They are mostly Chinese and Korean students. Gosh, no surprise there.

Bob · 2011-02-09 00:22:56

Oh wow. That took me by surprise. It's just gone midday here so I assumed you were in the UK too.

OK. I'll try to be quick in case you want to get to bed, Plus I'm at home at the moment but must go into school soon.

If you've got cm square paper, that'll be ideal.

Time across and distance up the side.

Have 1cm across for every 15 mins and I think this will fit going up, 1cm for each 5 km.

The first car's line will be four squares across for every 7 up.

The line represents the equation d = 35t.

Start the second car three squares across, ie at 45 mins. Go 4 squares across for every 10 squares up.

that equation is d = 50(t-0.75)

Keep going across and up with each line until they cross. That's when the cars are at the same distance at the same time.

You should get 2.5 hours.

If you want to stick with bobbym's equations you need to start with car 2 at the origin and have negative 45 mins for the first car. I think my version is easier for getting the class started on this.

The equations would then become d = 35(t+0.75) and d = 50t and the whole graph would shift 45 mins leftwards.

Bob

Last edited by Bob (2011-02-09 00:33:29)

bobbym · 2011-02-09 01:19:23

Hi Geoff;

Sorry, I had to leave but it seems that a certain person can no longer tie her own shoes without my assistance... I have been tying mine since I was 30 so I am good at it.

Geoff Wales wrote:

I will need to come up with a simpler solution if I am going to show Year 6 how to solve this.

Are your students up to the methods presented here? They are going to require a little algebra and/or a little coordinate geometry.

Geoff Wales · 2011-02-09 01:29:02

I will show them the graph, but I want to concentrate on how they can turn a worded question into an equation.

Thanks for your help guys. Much appreciated.

bobbym · 2011-02-09 01:39:10

Hi;

If they know d = r * t then you just have to show them why you set them equal to each other. Solving is just a bit of algebra.

Bob · 2011-02-09 07:57:05

hi Geoff Wales

It's difficult to advise exactly without knowing what resources you have and what the pupils already know.
I've put a more accurate copy of my graph on www.bundy.demon.co.uk .
Click the lessons link on the left and then 'Two Cars problem'.
This is an Excel file in the older xls format. Hope you can use it.

There's load you can do with this problem as a way of introducing algebra to the class.

eg. (i) After talking about how the distance of car 1 changes over time and what kph means, make a table of values for t and d.

(ii) Introduce the formula d = 35t.
Explain why the times sign is missing.

(iii) Draw the first line on the graph.

(iv) Ask for the formula for car 2. Most likely answer is d = 50t.
Go with it and make a second table. Draw second graph.
Ask why this doesn't represent the correct line for car 2.

(v) Ask how we can adapt this wrong line to get the right one.
Find the formula for this. Draw the correct line.

(vi) Find the crossing point. Why does this answer the question?

(vii) How do we know the answer is 2½ hours rather than, say, 2 hours and 29 minutes and 30 seconds.
For the graph isn't accurate enough to give the 'exact' answer.
Hopefully someone will suggest checking the numbers against the original question.

(viii) The graph is equivalent to solving the equation 35t = 50(t - 0.75)
(or you can work in minutes if you want to avoid the decimal)
35t = 50(t-45) which will give 150 minutes as the solution.

(ix) Show how to solve the equation.
Divide both sides by 5 to lower the numbers to 7t = 10(t-0.75)
Multiply out the bracket 7t = 10t - 7.5
Re-arrange the terms 7.5 = 10t - 7t
Simplify the right hand side 7.5 = 3t
Divide by 3 2.5 = t

If you want, it is very easy to make up new, similar questions.
Start with the graph and put on lines for two cars.
Translate this back into a problem.

Or even get the class to work in pairs, making such problems and solving each other's. That leaves you free to help those who need more help whilst the more able get on with it. Or pair an able, and less able, child so they get help that way.

Bob

Last edited by Bob (2011-02-09 08:00:07)

soroban · 2011-02-09 10:05:13

. .

.

bobbym · 2011-02-09 10:24:36

Hi soroban;

I use that way also. See post #8.

To expand on it it is as if the slower car is stationary and the faster car is travelling at 15 kph. The faster car only has to overcome the headstart.

Geoff Wales · 2011-02-09 23:12:26

bob bundy wrote:

hi Geoff Wales

I've put a more accurate copy of my graph on www.bundy.demon.co.uk .
Click the lessons link on the left and then 'Two Cars problem'.
This is an Excel file in the older xls format. Hope you can use it.
Bob

Thanks Bob and thanks Soroban. I played around with the file you gave me Bob, and jazzed it up a bit.

You may like it, or you could adapt it for your website. I put it here:

https://public.me.com/gwales

Last edited by Geoff Wales (2011-02-09 23:21:52)

Bob · 2011-02-10 01:58:59

hi Geoff Wales,

Nice treatment of the graph labelling Thanks.

Bad news on the algebra though, I'm afraid. You got to the right answer but with errors along the way.

Car A = 35t + 35 x ¾ (This cars time starts at t = zero so this should be Car A = 35t)
D1 = 35t + 35 x ¾

Car B = 50t - ¾ (You need to subtract the ¾ before times by 50 so Car B = 50(t - ¾)
D2 = 50t - ¾

D1 = D2
35t + 35 x 3/4 = 50t -3/4
15t - 3/4 = 35 x ¾

t - 3/4= 35/15 x ¾
(If you divide by 15, all terms must be changed so, if the above were correct, this would be t - 3/60= 35/15 x ¾)

t - 3/4= 7/3 x ¾
t = 10/4

Steps (viii) and (ix) in my previous post show how it should look.

Hope that helps,

Bob

Geoff Wales · 2011-02-10 02:18:33

I should have left the 3/4 hour out of it. That can be added at the end.

I only do Primary School maths, so errors like that come naturally to me

Bob · 2011-02-10 02:22:48

hi

So let your pupils teach you. Sometimes it's for the best!:):)

Bob

Math Is Fun Forum

#1 2011-02-08 17:11:55

When will Car B catch Car A after a delayed start?

#2 2011-02-08 17:20:12

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#3 2011-02-08 17:28:18

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#4 2011-02-08 17:30:36

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#5 2011-02-08 17:36:19

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#7 2011-02-08 17:46:02

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#9 2011-02-08 21:37:05

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#10 2011-02-09 00:01:29

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#19 2011-02-09 07:57:05

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#20 2011-02-09 10:05:13

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#21 2011-02-09 10:24:36

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#22 2011-02-09 23:12:26

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