Maths Help

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tejal

Maths Help

Post by tejal »

A father left home at 2-00am and travels @ 60miles per hour. Ten minutes later his son leave home and was travelling @80 miles per hour.

How far have the father gone at the moment his son begins his chase?

At what time will the son catch up with his father?
yoyo123
Posts: 8099
Joined: Mon Jun 18, 2007 3:32 pm
Location: East Kent

Post by yoyo123 »

1st part, if he is travelling at 60miles per hour,he travels a mile a minute..

therefore after 10 minutes, he has done 10 miles...


If his son travels at 80 miles an hour.

80 = 60 + 20 minutes

60mph = 1 mile per minute
20 mph = 1/3 mile per minute

80 mph = 1 1/3 miles per minute, so he travels 4 miles in 3 minutes.


I;'ll get back to you with teh answer once I've drawn a diagram!
KenR
Posts: 1506
Joined: Fri Mar 17, 2006 6:12 pm
Location: Birmingham

Post by KenR »

This is always an interesting one.

I always tend to use algebra and S=D/T

1st part is as described by yoyo123

For 2nd part:-

At the point they cross they will have travelled the same distance.

Hence, D = T x S

Therefore Tf x Sf = Ts x Ss

But because the son sets out 10 mins (or 10/60 hours) later, hence

Tf = (Ts + 10/60)

Therefore Sf x (Ts + 10/60) = Ss x Ts

And Sf = 60 and Ss = 80

Therefore 60Ts + 10 = 80Ts

or 20Ts = 10

Therefore Ts = 10/20 or 1/2 hour
Guest55
Posts: 16254
Joined: Mon Feb 12, 2007 2:21 pm

Post by Guest55 »

An easier way is to say -

In the ten minutes before the other starts the father goes 10 miles

Effectively the son is catching up the father at [80-60] = 20 mph


He has 10 miles to do - so thats half an hour - no need for algebra
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