Let t be the number of hours it takes Charlie to catch up with Bernie.
![\begin{gathered} \text{Speed}=\frac{dis\tan ce}{\text{time}} \\ \text{distance}=\text{speed}* time \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yrkbvlig38lceen62eg02w2eed09d9fz70.png)
Given:
For Charlie
time = t
speed = 20mph
![\begin{gathered} \text{Charlie's distance covered = sp}eed\text{ x time} \\ Charlie^(\prime)sdistance=20\text{ x t } \\ =20t\text{ miles} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w099xtyiafn03cofpz7wcphy731ejp1w7k.png)
For Bernie, he has started 1 hour earlier
Given:
time = t + 1
speed = 15mph
![\begin{gathered} \text{Bernie's distance covered will be;} \\ \text{speed x time = 15(t+1)} \\ =15t+15\text{ miles} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e6cq0bvfyhvauay16ydj8566prsfvk66am.png)
Hence, to get the time it will take for Charlie to catch up, we equate the distance both of them covered
![\begin{gathered} 15t+15=20t \\ \text{Collecting the like terms,} \\ 15=20t-15t \\ 15=5t \\ \text{Dividing both sides by 5,} \\ t=(15)/(5) \\ t=3\text{hours} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pusn6e4s1krozzejwpxiq2kyqk702ukt7m.png)
Therefore, it will take Charlie 3hours to catch up with Bernie.