Math Problem Statement

Rohan starts to walk 4 km to school at 7:30 am and her brother Eshaan starts at 8:15 am on his bicycle, travelling 12 km/hour faster than Rohan. They reach school at the same time. Find the speed of Rohan.

Solution

Let vv represent the speed of Rohan in km/h.

Step 1: Time taken by Rohan

Rohan starts at 7:30 am and travels 4 km to school. His time of travel is the difference between when he starts and when he arrives at school. Let tRt_R represent the time Rohan spends traveling.

The time taken by Rohan is: tR=4vt_R = \frac{4}{v}

Step 2: Time taken by Eshaan

Eshaan starts at 8:15 am, so he starts 45 minutes (or 0.75 hours) later than Rohan. Since both reach school at the same time, the time Eshaan spends traveling, tEt_E, is related to Rohan's time as: tE=tR0.75t_E = t_R - 0.75

Eshaan’s speed is v+12v + 12 km/h, since he is traveling 12 km/h faster than Rohan. Therefore, the time taken by Eshaan is: tE=4v+12t_E = \frac{4}{v + 12}

Step 3: Set up the equation

Since Eshaan’s time of travel tEt_E is tR0.75t_R - 0.75, we can equate the expressions for time: 4v+12=4v0.75\frac{4}{v + 12} = \frac{4}{v} - 0.75

Step 4: Solve the equation

First, clear the fractions by multiplying both sides by v(v+12)v(v + 12): 4v=4(v+12)0.75v(v+12)4v = 4(v + 12) - 0.75v(v + 12) Simplifying step-by-step: 4v=4v+480.75v(v+12)4v = 4v + 48 - 0.75v(v + 12) Canceling 4v4v from both sides: 0=480.75v(v+12)0 = 48 - 0.75v(v + 12) Expand the right-hand side: 0=480.75(v2+12v)0 = 48 - 0.75(v^2 + 12v) 0=480.75v29v0 = 48 - 0.75v^2 - 9v Rearranging the equation: 0.75v2+9v48=00.75v^2 + 9v - 48 = 0

Step 5: Solve the quadratic equation

Divide the equation by 0.75: v2+12v64=0v^2 + 12v - 64 = 0 Use the quadratic formula: v=12±1224(1)(64)2(1)v = \frac{-12 \pm \sqrt{12^2 - 4(1)(-64)}}{2(1)} v=12±144+2562v = \frac{-12 \pm \sqrt{144 + 256}}{2} v=12±4002v = \frac{-12 \pm \sqrt{400}}{2} v=12±202v = \frac{-12 \pm 20}{2} The two possible solutions are: v=12+202=4orv=12202=16v = \frac{-12 + 20}{2} = 4 \quad \text{or} \quad v = \frac{-12 - 20}{2} = -16 Since speed cannot be negative, we have v=4v = 4 km/h.

Conclusion:

Rohan's speed is 4 km/h.

Would you like a detailed explanation of any of these steps? Here are 5 related questions:

  1. How does the quadratic formula work in general?
  2. Can we solve this using a different method?
  3. Why did we discard the negative solution for speed?
  4. How would the time calculation change if Eshaan's speed were different?
  5. Can this problem be generalized for other distances or times?

Tip: For distance-rate-time problems, setting up a clear relationship between the two travelers' times helps simplify the problem.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Quadratic Equations

Formulas

Distance = Speed × Time
Quadratic Equation: ax^2 + bx + c = 0

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10