Relative Speed Problems

1. The Core Concept of Relative Speed

Imagine you are on a train. When a train on an adjacent track moves in the opposite direction, it seems to pass by incredibly fast. When it moves in the same direction, it seems to pass by much more slowly. This perceived speed is what we call "relative speed."

In TSD problems, we use relative speed to simplify a two-body problem into a one-body problem. We imagine one object is stationary and the other is moving at the "relative speed."

There are two fundamental cases:

1. Objects moving in opposite directions: The distance between them decreases at a rate equal to the sum of their individual speeds.

   • Relative Speed = $S_1+S_2$

2. Objects moving in the same direction: The distance between them decreases (if the faster object is behind) or increases (if the faster object is ahead) at a rate equal to the difference of their individual speeds.

   • Relative Speed = $|S_1-S_2|$

Let's solidify this with a short reading.

2. Application 1: Problems on Trains

Train problems are a classic application of relative speed. The key difference here is that trains have significant length, which must be accounted for in the "distance" part of our D = S × T formula.

The total distance to be covered is the sum of the lengths of all objects involved in the crossing.

Here are the common scenarios:

• Train crossing a stationary point object (pole, man):
  • Distance = Length of the Train ($L_T$)
• Train crossing a stationary object of length (platform, bridge):
  • Distance = Length of the Train ($L_T$) + Length of the Platform ($L_P$)
• Two trains crossing each other:
  • Distance = Length of Train 1 ($L_1$) + Length of Train 2 ($L_2$)
  • The speed used will be the relative speed ($S_1+S_2$ for opposite direction, $|S_1-S_2|$ for same direction).

Let's see how this is applied in a CAT-level problem.

The approach in Question 6 is very common. Often, one object starts earlier. You must first calculate how much distance it covers alone. The remaining distance is then covered at the relative speed once the second object starts moving.

3. Application 2: Boats and Streams

The second classic application is "Boats and Streams." This is just another framing of relative speed. The "stream" acts as one of the moving bodies.

Here's the essential terminology:

• Speed of the boat in still water (u): The boat's own speed without any current.
• Speed of the stream (v): The speed of the water current.
• Downstream: The boat travels with the current. The stream helps the boat, so their speeds add up.
• Upstream: The boat travels against the current. The stream hinders the boat, so the stream's speed is subtracted.

This gives us our new effective speeds:

• Downstream Speed ($S_D$) = u + v
• Upstream Speed ($S_U$) = u - v

The image below provides a great visual summary of this concept.

Let's dive into the core formulas associated with this topic.

From the reading, you likely noted these two extremely useful formulas. If you know the downstream and upstream speeds, you can quickly find the boat's own speed and the stream's speed:

• Speed in Still Water (u) = $\frac{1}{2}$ (Downstream Speed + Upstream Speed)
• Speed of Stream (v) = $\frac{1}{2}$ (Downstream Speed – Upstream Speed)

Let's see this in action.

Conclusion

Today we've added a powerful tool to your TSD problem-solving kit. By reframing problems with two moving objects in terms of a single relative speed, you can simplify them significantly.

Key Takeaways:

• Relative Speed Formulas:
  • Opposite Direction: $S_{rel}=S_1+S_2$
  • Same Direction: $S_{rel}=|S_1-S_2|$
• Train Problems: The "distance" to be covered is the sum of the lengths of the train(s) and any platform/bridge involved.
• Boats and Streams: This is a special case of relative speed.
  • Downstream Speed: $S_D=u+v$ (boat speed + stream speed)
  • Upstream Speed: $S_U=u-v$ (boat speed - stream speed)
• Key Boat/Stream Shortcuts: You can find the individual speeds if you know the combined speeds:
  • $u=\frac{S_D+S_U}{2}$
  • $v=\frac{S_D-S_U}{2}$

In our next lesson, we will shift gears slightly to another critical topic based on rates: Time & Work. You will see that the core logic of Work = Rate × Time is very similar to Distance = Speed × Time, allowing you to apply similar thinking to a new set of problems.