maths question help?!

[LJones]

Hey guys,
Just went through an exam for entrance for an aviation school, passed but i missed out on one question. Can some one tell me how to do this?

If one car starts moving at part A at 2:00pm travelling 90km/h in a constant direction, and another car leaves at same origin travelling in same direction at 2:10pm travelling 100km/h, when will they meet?

I remember doing this at school and i'm pissed off that i dont remember.

Cheers,

Lloyd

[Andy_Darlz]

I would think of this in terms of distance / time graph

As we know equations of straight line Y=Mx+C
M = Grad = Speed. C = Y intercept (x=0)

For car 1

Y = 90x + 0


For car 2

Y = 100x + C



0 = 100(10/60)+C


C= -100(10/60) = -16.666

Y=100x – 16.666

To find when they meet.

90x + 0 = 100x – 16.666

16.666/10 = x

x = 1.66666 = 1hr 40mins = 15:40

----------------------------------

Dist would be 1.666 * 90 = 150km
chk 1.666 * 100 - 16.666 = 150


Hope that helps

Andy D.

[Persephone]

Answer, although I don't know why it's Minus

Distance = Speed x Time

Time = Distance / Speed



Velocity A = Speed A / Time A
Velocity B = Speed B / Time B

You want the two speeds to equal.

S A = V A x T A
S B = V B x T B

V A x T A = V B x (T A + 10 Mins)

90 x T A = 100 (T A + 1/6)

90 T A = 100 T A + 100/6

-10 TA = 100/6

TA = -10/6 (10 / 6 * 60(mins) = 100 Mins

Q.E.D 100 Mins

[LJones]

cheers.
thanks...it just looks too easy when it spread out like that. im kicking myself for not getting it.

cheers again.

lloyd

[Mark 1]

How about this way of thinking about it:

At 2.10 the first car has been going 1/6 of an hour so has travelled 15km (90/6).
The second car is then closing the gap at the difference between their speeds i.e. 10km/h.

So to close a 15km gap at a rate of 10km/h takes an hour and a half.
Add the ten minutes the second car was waiting you get the 1h40m previously quoted.

So the answer is 3.40

[lowbypass]

Interesting one!
Distance, d
Speed, V
Time, t
Let us call meeting point, X
d = Vt
1. X = 90t
2. X = 100( t - 10)
But 1=2
90t = 100( t - 10)
90t = 100t - 1000
- 10t = - 1000
t = 100
t = 100 minutes
t = 100/60
t = 1.666 hrs
meeting point, X
substituting to either 1 or 2 since they are equal, taking 1.
X = Vt
X = 90 x 1.666
= 149.94
approx X = 150 km.

[LJones]

now i feel like a dumbass :P

[4KBeta]

Well for the aid of myself and others, has anyone got any links to revision guide for these types of Distance Speed Time calculations ?

[Diaz]

You could try

http://www.antonine-education.co.uk/..._1/topic_1.htm

Yes, it's AS revision, but it does the stuff. Could look at most AS revision guides for the motion sections (module 2)

[coodem]

The PET delayed departure formula works as well

[Mikehotel152]

Interesting how different people think about the same problem!

The way that Andy, Persephone and lowbypass looked at the question would be alien to me. I can understand the method but my brain avoids proper mathematics like the plague.

Now Mark has my vote. I think the way he thinks.

[CY333]

car A 90 kms per hour and sets off at 2.00
car B 100 kms per hour and sets off at 2.10.
At 2.10 the A car has done allready 15 kms.
relative speed between the two cars is 10 kms.
So it will take car B 1 hour and 30 to catch up to car A and cover the 15 kms.
In one and a hals hour car B has covers 150 km.
Car A 135 plus the 10 minuttes difference between the two starting hours that equals 15 kms and adds up to 150 for car A also.
So they will meet at 15.40 at 150 kms

[4KBeta]

thanks for that link and how people break it down. Not exactly hard just after more examples for practice

Any other examples people have, by all means please post up or chuck me links to whatever documents you have with questions

Many thanks

[Nathan.Hunter]

if you one where maths equations confuse you (which they shouldnt)

then in leymans terms

the 1st car has been going for 10 mins at 90km/h,

10 mins is 1 sixth on an hour, and in 1 hour you would travel 90km at 90km/h

so 1 sixth of 90 is 15km


you closing velocity is 10km/h

so it would take 1 and a half hours to catch up

so 1 and a half hours from 2.10 is 3.40

[Rugbyears]

Lowbypass – that solution terrifies me!!!!! I’m afraid one is rather a little slow in the brain department!!

Nathan.Hunter – that’s more like it, nice and simple without such need to complicate things!!!

[4KBeta]

Found it very hard to find more examples ...anyone got any good sources?

[degothia]

"and another car leaves at same origin travelling in same direction at 2:10pm travelling 100km/h, when will they meet?"

I dont think they will ever meet since they are going in the same direction...

D.

[degothia]

Im sorry for offending you Barke, it was not my intention at all. To make you feel better about it you could think about my post as a bad atempt for humor. You see, when I did my ATPLs I was tought to pay close atention to the words in the question. Not having english as my first language meet, to me, means coming face to face from different points of origin. But as you say meeting also means coverging, coming together, collect in one place and point of gathering. So thank you for correcting me.
D.