Overtake Calculation for Santa and Chris

This problem involves understanding the concept of speed and relative speed. When two people are moving in the same direction, the faster person overtakes the slower person when they cover the same distance from the starting point.

Let's denote the distance from the house to the school as \(D\).

Santa's Speed

Santa takes 40 minutes to cover the distance \(D\). His speed (\(V_S\)) is calculated as distance divided by time.

\[V_S = \frac{D}{40} \text{ distance per minute}\]

Chris's Speed

Chris takes 30 minutes to cover the distance \(D\). His speed (\(V_C\)) is calculated as distance divided by time.

\[V_C = \frac{D}{30} \text{ distance per minute}\]

Since Chris takes less time, Chris is faster than Santa.

Head Start and Overtake Time

Santa starts 5 minutes earlier than Chris. This means when Chris starts walking, Santa has already been walking for 5 minutes and has covered some distance.

Distance covered by Santa in the first 5 minutes (before Chris starts):

\[\text{Distance} = \text{Speed} \times \text{Time}\]

\[\text{Distance Santa covered in 5 minutes} = V_S \times 5 = \frac{D}{40} \times 5 = \frac{5D}{40} = \frac{D}{8}\]

Let \(t\) be the time in minutes measured from the moment Chris starts walking. We want to find the value of \(t\) when Chris overtakes Santa.

At time \(t\) (after Chris started):

• Santa's total time walking = \(5 \text{ minutes (head start)} + t \text{ minutes}\)
• Distance covered by Santa at time \(t\) = \(V_S \times (5 + t) = \frac{D}{40} \times (5 + t)\)
• Chris's total time walking = \(t \text{ minutes}\)
• Distance covered by Chris at time \(t\) = \(V_C \times t = \frac{D}{30} \times t\)

Chris overtakes Santa when they have covered the same distance from the starting point.

\[\text{Distance covered by Chris} = \text{Distance covered by Santa}\]

\[\frac{D}{30} t = \frac{D}{40} (5 + t)\]

Solving for Time \(t\)

We can cancel \(D\) from both sides of the equation (since \(D\) is not zero):

\[\frac{1}{30} t = \frac{1}{40} (5 + t)\]

\[\frac{t}{30} = \frac{5}{40} + \frac{t}{40}\]

\[\frac{t}{30} = \frac{1}{8} + \frac{t}{40}\]

Now, let's isolate the terms with \(t\) on one side:

\[\frac{t}{30} - \frac{t}{40} = \frac{1}{8}\]

Find a common denominator for 30 and 40, which is 120.

\[\frac{4t}{120} - \frac{3t}{120} = \frac{1}{8}\]

\[\frac{4t - 3t}{120} = \frac{1}{8}\]

\[\frac{t}{120} = \frac{1}{8}\]

Multiply both sides by 120 to find \(t\):

\[t = \frac{120}{8}\]

\[t = 15\]

So, Chris would overtake Santa 15 minutes after Chris starts walking.

Verification

Let's check the distances covered at \(t=15\) minutes (after Chris starts):

• Santa's total time = 5 minutes + 15 minutes = 20 minutes
• Distance covered by Santa = \(V_S \times 20 = \frac{D}{40} \times 20 = \frac{20D}{40} = \frac{D}{2}\)
• Chris's total time = 15 minutes
• Distance covered by Chris = \(V_C \times 15 = \frac{D}{30} \times 15 = \frac{15D}{30} = \frac{D}{2}\)

At 15 minutes after Chris started, both have covered half the distance to school, confirming that Chris overtakes Santa at this point.

The final answer is 15 minutes.