The concept of Relative Speed has always been very crucial for the students preparing for CAT exam. “Relative” means “in comparison to”. Thus, the concept of relative Speed is used when two or more bodies moving with some Speeds are considered. To make things simpler, one body can be made stationary **(i.e. Speed =zero)** and take the Speed of the other body with respect to the stationary body, which is the sum of the Speeds if the bodies are moving in the opposite direction and the difference if they are moving in the same direction. This Speed of the moving body with respect to the stationary body is called the Relative Speed.

**Relative Speed of 2 bodies = Sum of their individual Speeds if they are moving in the opposite direction = Difference of their individual Speeds if they are moving in the same direction.**

So consider A and B starting at the same Time, traveling in opposite direction with Speed of 20 kmph and 30 kmph and they have to travel a Distance of 200 km. Now the concept of relative Speed can be used to find the Time of their meeting.

Two bodies moving towards each other are A and B with Speeds 20 kmph and 30 kmph. Now make body A stationary and take the Speed of B in reference to A, so relative Speed of B = sum of their respective Speeds = 20 + 30 = 50 kmph ( as they are traveling in opposite directions).

Now A can be assumed to be stationary and the total Distance has to be traveled by B which he will do with a Speed = 50 kmph. So he will meet A after 200/50 = 4 hrs.

**Illustrations for Relative Speed**

**Example 1:**

**The thief Bhagu Ram is spotted by the policeman Pakad Singh from a Distance of 200m. Once they see each other they start running. What is the Distance Bhagu Ram who is running at 5 kmph would have covered before being caught by Pakad Singh running at 7 kmph?**

**Solution:**

Here are two bodies moving at different Speeds. So assume Bhagu Ram to be stationary and take the Speed of Pakad Singh relative to Bhagu Ram. So, now once the concept of relative Speed is applied, it can be deduced that Bhagu Ram is standing and Pakad Singh who is at a Distance of 200 m is running at 7 – 5 = 2 kmph. To catch Bhagu Ram he needs to travel a Distance of 200 ms or 0.2 km. So Time taken = 0.2/2 = 0.1 hours or 6 minutes. Thus the Distance that is covered by Bhagu Ram before being caught is 0.5 km.

**Example 2:**

** A bird is sitting on a train A moving at a Speed of 40 kmph. It sees another train B at a Distance of 200 ms with Speed of 60 kmph coming from the opposite direction on the same rail track. It flies with an average Speed of 10kmph and sits on another train. Again immediately it flies back to the first train and again to the second train and so on. It does so before the two trains crash. What is the total Distance traveled by the bird?**

**Solution:**

Two trains are moving in opposite direction with Speed 40 kmph and 60 kmph. Make train B stationary and then take the Speed of train A relative to the train B. So relative Speed of train A = 40 + 60 = 100 kmph. Before it crashes into the stationary train B it has to travel a Distance of 200 m which it will do in (0.2/100 ) hours. During this duration, the average Speed of bird has been 10 kmph i.e. before the crash total Distance covered by the bird = 0.2 * 10/100 km = 20 meters.

**Some Specific Cases of Relative Speed**

**CASE 1 : Boats & Streams**

Let the Speed of boat in still water = x

Speed of stream = y

As it is previously discussed that in questions where two bodies with different Speeds are concerned, the concept of relative Speed should be used. Here the stream can be assumed to be the stationary body and the Speed of boat relative to the Speed of the stream can be taken. Using the concept of relative Speed, one can now have the situation where the stream is stationary and only the boat is moving at a Speed determined by the direction of the boat relative to the stream.

**Illustrations for Boats & Streams**

**Example 3:**

**Gita rows a boat at a speed of 15 kmph upstream and 20 kmph downstream. Find the Speed with which Gita rows the boat in still water and also find the Speed of the stream.**

**Solution:**

Given that upstream Speed = 15 kmph

Downstream Speed = 20 kmph

Speed of Gita in still water = x = ((a + b))/2= (20+15)/2 = 35/2

Speed of stream =y = ((a – b))/2= (20-15)/2 = 5/2.

**Example 4:**

**The Speed of Narmada river is 5 kmph. A stationary body is placed in the river. Find the Time taken by the floating body to reach a stone which is 10 km downstream from the point where it is now?**

**Solution:**

Speed=Distance/Time.

So, Time taken to reach 10 km = 10/5 = 2 hours.

**Example 5:**

**A man rows 135 km upstream in 2.7 hours. He rows the same Distance downstream in 2.5 hours. However, for downstream, he reduces his Speed by 9%, while Speed of the current gets reduced by 20%. Find the Speed of the man.**

(a) 52

(b) 55

(c) 50

(d) none of these

**Solution:**

x= The speed of boat and

y= The Speed of stream

Given that x-y= 135/2.7= 50 —–(1) & Given that 10/11 x + 4/5 y = 135/2.5 = 54 ——–(2)

(reduction of 9% = 100- 9 = 0.91x. 0.91 in terms of a fraction can be written as 10/11 since 1/11= 0.0909). (Similarly, reduction of 20% => 0.8 y. 0.8 in terms of a fraction can be written as 4/5.

**Go from answer options-**

Option (c)The Speed of the man cannot be 50, as the Speed of the stream cannot be zero (on substituting in equation (1))

option (b) substitute x= 52 and x=55 in the second equation and see for which value you are getting an integral value for y. This happens only for x=55, where you will get y=5. Hence answer is the option (b) Here the Reverse Gear approach is used i.e. working from answer options. It should be Remembered that longer the question, the easier it is to eliminate the wrong answer options.

**CASE 2: Moving body and a stationary body of negligible length**

Assume a stationary pole and a train approaching it. Here the Speed of the train relative to the pole will be the Speed of the train itself, as the pole has a Speed of zero. Also, since the pole is of negligible length in comparison to the train, the Distance traveled by train to pass the pole will be the length of the train itself. Therefore, the Time taken by a train with length L1 and Speed S1 to pass a post/ stationary pole = L_{1}/S_{1}

**Illustrations for ****Moving body and a stationary body of negligible length**

**Example 6:**

**A train is traveling at a Speed of 4 m/s. Raman is standing on the road. How much Time will it take for the train to cross him if the length of the train is 200m?**

**Solution:**

Time taken to pass Raman= (Length of train)/(Speed of train) = 200/4= 50 sec.

**CASE 3: Moving body and a stationary body of some length**

Now let’s assume that the same train is going to pass a long bridge of length L. The new Distance covered by the train while passing over the bridge will be the length of the bridge+ the length of the train. Also, the Speed of the train relative to the bridge will be the Speed of the train itself, as the bridge is stationary. The time taken by a train of length L_{1} and Speed S_{1} to pass a bridge of length L = (L_{1}+ L)/S_{1} Here the Distance to be covered is the sum of the lengths of train and bridge.

**CASE 4: Two moving bodies in opposite directions**

When two bodies move in opposite direction, then the Relative Speed = Sum of Speeds i.e for e.g. for a person sitting in a train moving with a Speed of 40 km/hr in the west direction, another train which is going towards east with a Speed of 40 km/hr, will appear to move at a Speed of (40+40) = 80 km/hr. There are two trains with lengths L_{1} and L_{2} and Speeds S_{1} and S_{2} respectively moving in the opposite direction. Time taken by the trains to cross each other after meeting = (L_{1}+L_{2})/(S_{1}+ S_{2} ).

Here the Total Distance to be covered is the sum of the lengths of the trains and relative Speed is the sum of the Speeds (since they are moving in the opposite direction).

**CASE 5: Two moving bodies in the same direction**

When two bodies move in the same direction then the Relative Speed = Difference of Speeds i.e for a person a sitting in a train moving with a Speed of 60 km/hr in the west direction, another train which is going towards west with a Speed of 40 km/hr, will appear to move at a Speed of (60-40) = 20 km/hr.

There are two trains of lengths L_{1} and L_{2} and traveling at Speeds S_{1} and S_{2} respectively moving in the same direction. Time taken by the trains to cross each other after meeting = (L_{1}+L_{2})/(S_{1} – S_{2} ).

Here the Total Distance to be covered is the sum of the lengths of the trains and relative Speed is the difference of the Speeds (since they are moving in the opposite direction).

**Example 7:**

**Train A which is 125 m long is traveling at 108 km/hr. On a parallel track, a 180 m long Train B is traveling at 72 km/hr, in the same direction. In Train B, a passenger is walking towards the rear end of the train at a Speed of 9 km/hr. In how many seconds will Train A completely cross the passenger in Train B? **

(a) 14.4

(b) 10

(c) 16.67

(d) 24

**Solution:**

Speed of A = 108 km/hr = 30 m/s

Speed of train B = 72 km/hr = 20 m/s

Here r the relative Speed between train A and train B should be considered first i.e.= 30 -20 = 10 m/s.

Relative Speed -Train A and man moving in the opposite direction with a Speed of 2.5 m/s=10 +2.5 = 12.5m/s

Time taken by train A to pass the man who is moving in the opposite direction = 125/12.5 = 10 secs since the Distance to be covered over here is the length of train A only.

**Example 8: How much Time a 100m long train traveling at a Speed of 8 m/s will take to overtake another train which is 80m and is traveling in the same direction at a Speed of 4 m/s?**

**Solution:**

Suppose the 80 m long train is stationary and take the Speed of the 100m long train in relative to the former. So relative Speed of 100m long train = 8 – 4 = 4 m/s. To overtake the 80m long train it has to travel a total Distance = sum of its own length and that of the other train = 100 m + 80 m = 180 m So to travel that Distance with a Speed of 4 m/s, it will take 180/4= 45 secs.

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