Math Problem Statement

Jake completes a journey in such a way that he starts at a certain initial speed, and then after every one hour, he triples his speed and reaches his destination at the scheduled time. If he completed his journey, travelling (entire trip) at his initial speed, he would have taken 10 times the scheduled time. Find the scheduled travelling time (in hours) if the total distance travelled is 320 km.

Solution

Let the scheduled travel time be denoted by TT (in hours), and let Jake's initial speed be denoted by vv (in km/h).

Step 1: Equation for journey if Jake travels at initial speed vv

If Jake had traveled the entire journey at his initial speed vv, he would have taken 10 times the scheduled time. Therefore, the total time taken would be 10T10T, and the total distance traveled is 320 km. Using the formula for distance: Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time} we have: 320=v×10T320 = v \times 10T Solving for vv: v=32010T=32Tv = \frac{320}{10T} = \frac{32}{T}

Step 2: Equation for journey based on tripling speed

Jake starts at speed vv and triples his speed every hour. The distance he travels in each hour can be calculated based on his changing speeds:

  • In the first hour, his speed is vv, so he covers v×1=vv \times 1 = v km.
  • In the second hour, his speed is 3v3v, so he covers 3v×1=3v3v \times 1 = 3v km.
  • In the third hour, his speed is 9v9v, so he covers 9v×1=9v9v \times 1 = 9v km, and so on.

Thus, the total distance Jake covers after TT hours is the sum of the distances in each hour: Total Distance=v+3v+9v++3T1v\text{Total Distance} = v + 3v + 9v + \cdots + 3^{T-1}v This is a geometric series with the first term vv and common ratio 3. The sum of the first TT terms of a geometric series is given by: S=v×3T131=v×3T12S = v \times \frac{3^T - 1}{3 - 1} = v \times \frac{3^T - 1}{2} Equating this to the total distance of 320 km: v×3T12=320v \times \frac{3^T - 1}{2} = 320

Step 3: Substitution of vv

From Step 1, we have v=32Tv = \frac{32}{T}. Substituting this into the equation above: 32T×3T12=320\frac{32}{T} \times \frac{3^T - 1}{2} = 320 Simplifying: 16(3T1)T=320\frac{16(3^T - 1)}{T} = 320 3T1T=20\frac{3^T - 1}{T} = 20 3T1=20T3^T - 1 = 20T 3T=20T+13^T = 20T + 1

Step 4: Solving the equation

To solve this equation, we can test small values of TT:

  • For T=4T = 4: 34=81,20×4+1=80+1=813^4 = 81, \quad 20 \times 4 + 1 = 80 + 1 = 81 Thus, T=4T = 4 is a solution.

Final Answer:

The scheduled travel time is 4\boxed{4} hours.

Do you want further clarification, or have any questions?

Here are 5 related questions:

  1. How does geometric series summation apply in travel distance problems?
  2. What would Jake's speed be during the final hour of the trip?
  3. Can you find another method to verify the solution?
  4. What happens if Jake doubles his speed instead of tripling it every hour?
  5. How would the problem change if the total distance was different?

Tip: In time-distance-speed problems involving changing speeds, the concept of geometric progressions can be very useful.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometric Progression
Distance-Speed-Time Relationship

Formulas

Distance = Speed × Time
Sum of a geometric series: S = a × (r^n - 1) / (r - 1)

Theorems

Geometric series summation

Suitable Grade Level

Grades 10-12