Minus' Solution to

Go, Spot, Go!
Posted May 19 - 25, 1997

What you need to determine is how long it takes Dick and Jane to meet. Dick & Jane will travel a total of 2 m/s + 1 m/s = 3 m/s. Since the length of the hall is 90 m, it will take 90 m ÷ 3 m/s = 30 seconds for them to meet.

The dog travels at 3 meters per second for 30 seconds, so Spot travels 3 m/s x 30 s = 90 meters running back and forth between Dick and Jane.

Your Solutions

Nizzavich Chvasky whipped up this fresh and tasty starter for our math shark. Bright, crisp and clean. Just what our fat ole mascot needs!

If Dick walks double as fast as Jane, then he is going to walk double as far as her before he meets her. So if they start 90 metres away, Dick must walk 60 metres and Jane 30 metres.

For Dick to walk 60 metres at 2m/sec and Jane 30 metres at 1m/sec, they must've walked for 30 seconds. Spot can run at 3m/sec, so over 30 seconds at 3m/sec he would be able run 90m.

The answer is Spot runs ninety metres before Dick and Jane meet.

Mr. T. is flying solo now that his Little Einsteins are out of school. But he is proving his merit as a gourmet chef by cooking up this scrump-delicious dish!

(By the way, I have given several variations of this problem to my classes throughout the year. I can't even fool them anymore... which of course, is the idea!!)

During each second, Dick & Jane will travel a total of 3 m/s. Since the total distance of the hall is 90 m, it will take 30 s for them to meet.

As for Spot, he is traveling at a furious pace of 3 m/s for those same 30 s, so he will cover a grand total of 90 m, which by sheer coincidence is the same distance of the hall. Obviously, had Spot's rate not been equal to the sum of Dick & Jane's rates, his distance covered would have been different.

Martin [mai@fir.Scotch.wa.edu.au], one of our international chefs from down under, delighted Minus with his logical twist on the solution. Hi, Martin!

Jane and Dick are 90 meters apart. Dick walks twice as fast as Jane, so Jane would have walked one third of the distance (30 m) by the time the two met. Since she Walks 1m per sec, the pair meet after 30 sec.

Spot would have been running around for thirty seconds at 3m/sec. So Spot would have run 30*3 metres - 90 metres.

From Martin of Western Australia. Hi everyone!

Ben Dessauvagie also used Jane as his "timer" so to speak. An interesting accompaniment to our mascot's feeding frenzy.

Spot travels 3 metres per second, this is the same distance as Dick and Jane travel in one second put together. Dick travels 60 metres in 30 seconds and Jane travels 30 metres in 30 seconds. Spot travels 3 times as fast as Jane so if Jane travels 30 metres before meeting with Dick, the dog must run 90 metres.

Brendan Bensky [russellb@ois.net.au] cooked up a nutritious dish with all the right ingredients. 

If Jane and Dick were 90 metres apart then after 30 seconds Dick had walked 60 metres and Jane 30. At that stage they were together. This had taken them 30 seconds. Spot the dog was running 3 metres a second so at the point when Dick and Jane reached each other Spot had ran 90 metres.

Uma Gummadavelli included her recipe for success. Now all of us know how to create Uma's perfect entree.

The time it took for Dick and Jane to meet was 30 seconds because the distance between them was decreasing at the rate of 3 meters per second. The distance between them was 90 meters so Dick and Jane would meet after 90/3 = 30 seconds. The dog Spot was running at 3 meters per second. It was running for 30 seconds. So the distance traveled by Spot was 30 seconds x 3 meters per second = 90 meters.

Distance between Dick and Jane = 90 meters

Speed at which the distance    = 3  meters per second 
decreases between them

Time taken by Dick and Jane to = 90/3 seconds
meet each other
                               = 30 seconds

Spot's speed                   = 3 meters per second

Distance traveled by Spot in   = 3 x 30 meters
30 seconds

                               = 90 meters

Brad & Heather Felix used the classic distance formula for their tempting morsel.
Let x= time Dick and Jane are walking.
Using D = r * t, Dick has gone a distance of 2x and Jane 1x.
Therefore, 2x + 1x = 90 meters. Thus, x = 30 seconds.
So, if they have been walking for 30 seconds, then Spot has been running 3 meters/second for 30 seconds which is a distance of 90 meters.

These three GED students at Bushwick High, in Brooklyn, New York, didn't sweat over this week's problem. They used good logic and the distance formula. Good team work, gang!

Hi Minus, second week in a row on time before the 12 noon deadline. Yes!!

Dick and Jane were 90 meters apart. They started walking towards each other. Dick is walking 2 meters per second and Jane is walking 1 meter per second. One plus two equals three meters per second combined rate.

Since distance equals rate x time, then time equals distance divided by the rate. Ninety meters divided by three meters per second is 30 seconds. All this time Spot is traveling at 3 meters per second.

Therefore, Spot's distance is 30 seconds times 3 meters per second or 90 meters.

Jacqueline Matos, Wilfredo Soto and Cristobal Florencio

Once again, Minus' dinner menu was full of great dishes. Unfortunately, he ran out of room to feature all of the flavourful side dishes he received. Thanks also goes out to...

Sam Thomas from Discovery School
Andrew MacKinlay [andmak@cygnus.uwa.edu.au]
GUY ("Nerdoligest" by trade)
Chris Donisi [mai@fir.scotch.wa.edu.au]
Giles Bailey [96baileyg@swanbourne.wa.edu.au]
Jai Seeber [96seeberj@swanbourne.wa.edu.au]
Michael Downing [downingm@swanbourne.wa.edu.au]

Minus has developed quite an
appetite for Australian cuisine.
Well, he is a Great White after all.