(Follow-up) June Vacation Chap 9.2 RATE (L5) Can he catch up with her?

What is the 'slowest speed' that Shamus has to ride the bicycle so that he could catch up with Grace?

Note: The units we are dealing here are: Time in minutes & Distance in metres.
Speed therefore is metres per minute (m/min). We'll have to convert it to something that is more common (i.e. m/s or km/h)

By answering to the pointers in the original question...
  • What information do you know from the problem?
  • What else do you need to know to solve the problem?
  • Pick a reasonable number for the information you need. 
1. Information we know:
  • The ratio of speed (one is 3 times the other)
  • Maximum distance covered: 800 m
  • One started the journey 10 minutes later
2. To solve the problem, we need to know the speed or time taken by Grace or Shamus. Since we do not know these information, we could only provide answers for certain scenario to 'happen', for example, Shamus catches up with Grace at the school gate, when both have travelled 800m. We can also test another extreme whereby Shamus to catch up with Grace along the way to school.

3. By using the info given, we can solve the problem using Algebra:
Let Grace's speed be x m/min
Shamus' speed is therefore 3x m/min

Since distance travelled = 800m (imagine that Shamus catches up with Grace at the school gate)
Time taken by Grace = 800/x min
Time taken by Shamus = 800/(3x) min

Since the time difference is 10 min (as Grace started earlier)
Time taken by Grace - Time taken by Shamus = 10
800/x - 800/(3x) = 10
by solving the algebraic equation, we have x = 53 1/3 m/min
To convert it to m/s, we have x = 8/9 m/min which is the same as 0.889 m/s (correct to 3SF)

Shamus takes 5 min to catch up with Grace
Therefore, time they meet = 6.45 am

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