Let \(D\) be the total distance of the route in kilometers, and \(T\) be the normal time to reach the destination in hours.
Step 1: Time taken on the first day
On the first day, the bus travels at 60 km/h and arrives 3.5 hours later than usual. The time taken on the first day is \(T + 3.5\) hours.
\[\text{Time taken on day 1} = T + 3.5.\]
Since distance = speed × time, and the distance traveled is \(D\) at 60 km/h, we have:
\[\text{Time taken on day 1} = \frac{D}{60}.\]
Equating these two expressions for the time taken on the first day:
\[\frac{D}{60} = T + 3.5. \quad \text{(1)}\]
Step 2: Time taken on the second day
On the second day, the bus covers \(\frac{2}{3}\) of the route in \(\frac{1}{3}\) of its normal time, which is \(\frac{1}{3}T\).
\[\frac{1}{3}T.\]
The remaining \(\frac{1}{3}\) of the route is covered in 40 minutes, or \(\frac{2}{3}\) hours.
Therefore, the total time taken on the second day is:
\[\text{Time taken on day 2} = \frac{1}{3}T + \frac{2}{3}.\]
The total distance \(D\) is covered on the second day at 60 km/h, so the time taken is also \(\frac{D}{60}\).
\[\text{Time taken on day 2} = \frac{D}{60}.\]
Equating these two expressions for the time taken on the second day:
\[\frac{D}{60} = \frac{1}{3}T + \frac{2}{3}. \quad \text{(2)}\]
Step 3: Solve the system of equations
Solve equations (1) and (2) simultaneously.
From equation (1):
\[\frac{D}{60} = T + 3.5,\]
\[D = 60(T + 3.5). \quad \text{(3)}\]
From equation (2):
\[\frac{D}{60} = \frac{1}{3}T + \frac{2}{3},\]
\[D = 60\left(\frac{1}{3}T + \frac{2}{3}\right),\]
\[D = 20T + 40. \quad \text{(4)}\]
Step 4: Set equations (3) and (4) equal
Equate the expressions for \(D\) from equations (3) and (4):
\[60(T + 3.5) = 20T + 40.\]
Simplify the equation:
\[60T + 210 = 20T + 40,\]
\[60T - 20T = 40 - 210,\]
\[40T = -170,\]
\[T = \frac{-170}{40} = 4.25 \, \text{hours}.\]
Step 5: Find the normal arrival time
The normal time \(T = 4.25\) hours, which is 4 hours and 15 minutes. If the bus departs at 9:00 am, the normal arrival time is:
\[9:00 \, \text{am} + 4 \, \text{hours} \, 15 \, \text{minutes} = 1:15 \, \text{pm}.\]
Final Answer
The bus normally reaches its destination at:
\[\boxed{1:15 \, \text{pm}}.\]