Step 1: Understanding the Concept:
When a train passes a pole (or a stationary point object), the distance it covers is equal to its own length. The relationship between speed, distance, and time is fundamental to solving this problem.
Step 2: Key Formula or Approach:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
Here, Distance = Length of the train.
Step 3: Detailed Explanation:
Let \(L_P\) and \(S_P\) be the length and speed of train P.
Let \(L_Q\) and \(S_Q\) be the length and speed of train Q.
For train P:
Time taken (\(T_P\)) = 40 seconds.
Distance covered = \(L_P\).
\[ S_P = \frac{L_P}{T_P} = \frac{L_P}{40} \quad \text{(Equation 1)} \]
For train Q:
Time taken (\(T_Q\)) = 2 minutes 20 seconds = \(2 \times 60 + 20 = 120 + 20 = 140\) seconds.
Distance covered = \(L_Q\).
\[ S_Q = \frac{L_Q}{T_Q} = \frac{L_Q}{140} \quad \text{(Equation 2)} \]
Given relationship between lengths:
The length of train P is two-thirds that of train Q.
\[ L_P = \frac{2}{3} L_Q \]
Step 4: Final Answer:
We need to find the ratio of the speeds, \(S_P : S_Q\) or \( \frac{S_P}{S_Q} \).
Using Equation 1 and Equation 2:
\[ \frac{S_P}{S_Q} = \frac{L_P/40}{L_Q/140} = \frac{L_P}{L_Q} \times \frac{140}{40} \]
Now, substitute the relationship \(L_P = \frac{2}{3} L_Q\):
\[ \frac{S_P}{S_Q} = \left(\frac{2/3 \cdot L_Q}{L_Q}\right) \times \frac{140}{40} \]
\[ \frac{S_P}{S_Q} = \frac{2}{3} \times \frac{14}{4} = \frac{2}{3} \times \frac{7}{2} \]
Cancel out the 2s:
\[ \frac{S_P}{S_Q} = \frac{7}{3} \]
The ratio of the speed of train P to that of train Q is 7:3.