Answer:
it will take Daniel approximately \( 0.8108 \) hours to catch up with Frank, or approximately \( 48.65 \) minutes.
Step-by-step explanation:
To solve this problem, we'll first find out at what point Daniel starts to catch up with Frank, and then calculate the time it takes for Daniel to catch up.
Let's denote:
- \( d \) as the total distance Frank travels,
- \( t_1 \) as the time it takes for Frank to travel \( \frac{1}{3} \) of the distance, and
- \( t_2 \) as the time it takes for Daniel to catch up with Frank.
We know that Frank travels at a speed of \( 45 \) mph and Daniel travels at \( 50 \) mph.
Given that Frank travels at a constant speed, the distance he travels in \( t_1 \) hours is:
\[ \text{Distance}_\text{Frank} = \text{Speed}_\text{Frank} \times t_1 \]
\[ \text{Distance}_\text{Frank} = 45 \times t_1 \]
Since Frank has traveled \( \frac{1}{3} \) of the total distance when Daniel starts, we have:
\[ \frac{1}{3}d = 45 \times t_1 \]
Now, we can solve for \( t_1 \):
\[ t_1 = \frac{\frac{1}{3}d}{45} \]
\[ t_1 = \frac{d}{135} \]
Now, we know that Daniel's time, \( t_2 \), must be equal to Frank's time plus the additional time it takes Daniel to catch up with Frank. So:
\[ t_2 = t_1 + \frac{d}{50} \]
Substituting the expression for \( t_1 \):
\[ t_2 = \frac{d}{135} + \frac{d}{50} \]
Now, let's find a common denominator:
\[ t_2 = \frac{10d}{1350} + \frac{27d}{1350} \]
\[ t_2 = \frac{37d}{1350} \]
To find \( d \), we know that when Daniel catches up with Frank, they will have traveled the same distance. So, \( d = 45t_2 \).
Substituting \( d \) into the expression for \( t_2 \):
\[ t_2 = \frac{37 \times 45t_2}{1350} \]
Now, solve for \( t_2 \):
\[ 1 = \frac{37 \times 45}{1350} \]
\[ 1 = \frac{1665}{1350} \]
Now, let's solve for \( t_2 \):
\[ t_2 = \frac{1350}{1665} \]
\[ t_2 \approx 0.8108 \]
So, it will take Daniel approximately \( 0.8108 \) hours to catch up with Frank, or approximately \( 48.65 \) minutes.
