**GETTING STARTED**

**☛ Speed = Distance / Time**

Speed is generally expressed in terms of km/hr or m/s

**☛ 1 km/hr= 1000m/3600s ****= 5/18**** m/s**

Thus, in order to **convert km/hr to m/s** **multiply it by 5/18**

Similarly, in order to **convert m/s to km/hr**, **multiply by 18/5**

**Eg. Let us say, a bus is travelling at 54 km/hr. What will be the speed of bus in m/s?**

Multiplying 54 by 5/18, we get, **5/18 x 54 = 15 m/s** .The bus is travelling at 15 m/s.

**Eg. Let us consider a train plying from Mumbai to Pune at 60 km/hr. The distance between the two cities is 180km. One day, due to signaling problems, average speed of train was 45km/hr. How early or late will the train reach its destination?**

**Soln.**

Since the average speed of the train has decreased, the time taken to reach its destination has to increase. So the train will be obviously late.

Let us find out how late the train is:

Time=Distance/Speed.

Usual time taken=180/60=3hrs

Time taken on that particular day=180/45=4hrs.

It takes the train one 1hr more to reach its destination.

☛ If **speed** of an object **increases** by **y %**, **time taken** to cover same distance will **decrease** by **100y/(100+y) %**.

**EXPLANATION**

Let the original speed be S1 and the initial time taken be T1.

**Increase in speed = y % of S1 **

**= (S1. Y)/100**

New speed, say **S2 = S1 + Increase in speed **

**= S1 + (S1. Y)/100 **

**= S1. (1+ (y/100)) **

**= S1. (100 + y)/100**

With increase in speed, the **time taken to cover the distance has to decrease.**

Since we know, distance in both cases is same,

S1. T1 = S2. T2

S1. T1 = S1. (100 + y)/100 . T2

T2 = 100/(100+y) . T1

**Decrease in Time **

= T_{1 }– 100/(100+y) . T_{1}

**= y/(100+y) . T _{1}**

_{ }

**% Decrease in time **

= [(y/(100+y) . T_{1})/ T_{1}] x 100

**= 100y / (100+y) _{ }**

☛ Similarly, if **speed** of an object **decreases** by **y %**, the **time taken** will **increase** by **100y/(100-y) %**.

**Eg. If a man travels at 30 km/h, he reaches his destination late by 10 min but if he travels at 42 km/h, then he reaches 10 min earlier. Therefore, the distance traveled by him is:**

**Soln:**

Let the distance between two places be x and takes y min to reach his destination.

Since the distance is same in both conditions, we equate both the cases.

Speed_{1 }x Time_{1} = Speed_{2 }x Time_{2}

30 x (y+10) = 42 x (y-10)

30y + 300 = 42 y – 420

12 y = 720

Thus, y= 60 minutes

If he travels at 30 kmph he covers the distance in 70 minutes.

Therefore, distance = 30 x (70/60) = **35 kms.**

** PROBLEMS ON TRAINS**

**☛ CASE -I : Train overtaking stationary object **

Consider a train overtaking a man standing on a platform.

The distance train has to cover to fully cross the man is **equal to the length of the train.**

As the man is stationary, **time taken** to cross the man completely

**= Length of train /Speed of the train**

**☛CASE – II: Man running in same direction as train**

Distance to cross the man completely = Length of the train

Relative Speed of train w.r.t man =Speed of train – speed of man

**Time taken to cross the man = **

**Length of train/ (Speed of train – speed of man)**

**☛ CASE -III: Man Running opposite to train**

** **

Distance to cross the man completely = Length of the train

Relative Speed of train w.r.t man =Speed of train + speed of man

**Time taken to cross the man =**

**Length of train/ (Speed of train + speed of man)**

**☛ CASE IV : Crossing stationary objects like platform/bridge/tunnel**

** **

**Distance to completely cross = K+L**

Since the other object is stationary, speed of platform = 0

**Time taken for train to cross completely = (Length of train + Length of platform)/Speed of train**

**☛ CASE V: ****Trains moving in opposite direction**

**Distance = Sum of the length of both the trains = L + M**

Since, moving in opposite direction,

Speed = u+v

**Time taken to cover the distance = (L+M)/(u+v)**

**☛ CASE-VI: Trains moving in same direction**

**Distance = L+M**

Since, moving in same direction,

Speed = u-v

**Time taken to cross each other **

**= (L+M)/(u-v)**

** Boats & Streams/ Aeroplane & Wind**

☛ Streams oppose the movement of boats if they flow against the direction in which boat moves

☛ At the same time, if the stream flows in the same direction of the boat, speed of boat increases

☛ Let the **speed of boat in still water be u km/hr and speed of stream be v km/hr**

☛ The speed of boat in **upstream (stream opposite to boat) = u-v km/hr**

☛ The speed of boat in **downstream (stream same direction as boat) = u+v km/hr**

**Eg. A boat travels 30 km upstream in 6 hours and returns back in 4 hours. Find the speed of boat in still water and the stream.**

**Soln:**.

Let speed of boat in still water be u km/hr

Let speed of stream be v km/hr** **

**Speed of boat in upstream **

**= u – v = 30/6 **

**= 5 km/hr**

** Speed of boat in downstream **

**= u + v = 30/4 **

**= 7.5 km/hr**

Solving, we get

**Speed of boat in still water (u) **

**= 6.25 km/hr**

** Speed of stream (v) = 1.25 km/hr**

**Races & Games**

☛ These problems involve two or more competing players who outscores the other

**Eg. A and B run a 100 m race from a same start point. A has a speed of 10 km/hr whereas B has a speed of 8 km/hr. By what distance will defeat B?**

**Soln:**

As we can see, **A is faster than B** and both start at same point and time. Hence **A will obviously win the race.**

To find the distance by which A defeats B, let us understand the following diagram

When A reaches end point, B would be some distance away from the end point. Let the **distance be ‘x’**.

A runs at a speed of 10 kmph =10 x 5/18

** = 50/18 m/s**

As we know, time = distance /speed

**Time taken by A to complete race = 100/(5/18) = 36 sec**

When A reaches the end point, B covers,

**36 x 5/18 x 8 km/hr = 80 m**

Thus **A beats B by 20 m**

**B will complete** **the race** in 100/(5/18 x 8) **=45 sec**

Therefore, we can also say that, **A defeats B by 9sec**

**Distance by which A defeated B = Time by which A defeated B x Speed of B (loser)**

20 m = 9 sec x 5/18 x 8 km/h

**Giving a Start**

**☛ In the above race, if B had started 20 m ahead of A, both would have reached the end point at the same time**.

In 36 sec, A would have covered 100 m whereas B would have covered only 80 m.

But thanks to the start, both will reach at the same time.

This is known as ‘**giving a start of 20m’**

**☛ If A had given B a start of say 40 m, B has to cover only 60m to finish the race while A has to cover 100m.**

** **

B covers 60m in 60/ (5/18)x8

= 27 sec.

When B covers, 60 m in 27 sec, while A will only cover

5/18 x 10 x 27 = 75 m.

In this case, B will beat A by 25 m or 9 sec. [As A will reach end point in 36 sec].