### SPEED, TIME AND DISTANCE capsule

**Did you ever wonder how fast you run?**

**I never did!**

**But , seriously the way the speeds and distances work , can play a much vital role in daily life,**

**suppose you depart from a station and you know the speed of the bus you are in and want to calculate if with the present speed , will you be able to get to your destination in time....**

**So, lets see some of these concepts,**

## RELATION BETWEEN SPEED, TIME AND DISTANCE:

SPEED = DISTANCE / TIME

, Where speed can have any measuring unit like it could be in metre per sec ,or, metre per hour ,or, kilometre per hour , or any thing , and same applies to time and distance too , there units also can have different measuring units .

so first of all lets see what is a unit ,

so unit is the way how you measure a quantity like and all the units have some reference associated to each other , below is the reference chart of units ,

International System Of Units |

SI UNIT CONVERSION FACTOR |

DISTANCE:

This is the length of the path that has been travelled by any object or a person between any two sites . The unit of distance is m or km.

SPEED:

The distance travelled by any object or a person in some unit time . It's unit is m/s or km/h .

**Here , are a bunch of formulas coming ,**

**Tip : note down the formulas atleast 3 times**

### SOME FORMULAE:

- SPEED = DISTANCE / TIME
- AVERAGE SPEED = TOTAL DISTANCE TRAVELLED / TOTAL TIME TAKEN
- If A travels having a speed of x km/h for t1 h and with speed y km/h for next t2 h , then AVERAGE SPEED = ( x*t1 + y*t2) / (t1+t2)
- If you want to convert the from m/s to km/h or from km/h to m/s, m/s -> km/h (multiply the speed by 5/18) , km/h -> m/s (multiply the speed with 18/5)
- When a man / vehicle / object covers two equal distances with the speed of x km/h and y km/h , then the average speed of the man/ object for the whole journey will be , (2*x*y) / (x+y)
- When an object changes its speed with in the ratio of a:b, then the ratio of time taken becomes b:a
- If two objects are moving in the same direction with speeds x and y km/h(where , x>y) starting from the same point and going into the same direction , then their relative speed will be (x-y) km/h. Whereas if they move in different directions , then their relative speed will be (x+y)
- If an object changes its speed to (x/y) of its normal speed , but gets late by t min, then the normal time taken by it , = (t*x) / (y-x), whereas if that object changes its speed to (x/y) of its normal speed and reaches early by t min , then normal time taken by him, = (t*x) / (x-y)
- If an object travels from A to B with a speed of x km/h and reached t1 ,time after some fixed time and when it propagates with a speed of y km/h from A to B , it reaches its destination t2 time before fixed time , then the distance between A and B = X*Y*(t1+t2) / (Y-X)
- If any object travels form A to B with a speed of x km/h and reaches t1 time late. Later it has increased its speed by y km/h to cover the same distance , then it will still get late by t2 time , then the distance between A and B will be , = (T1-T2)*(X+Y)*X/Y
- Suppose their are two objects that start at the same time from points A and B towards each other and after crossing each other , they will take t1 and t2 time in getting to the points B and A , respectively . Then , Speed of A: Speed of B = √t2 : √t1
- Suppose the ratio of the speeds of A and B is x:y , then the ratio of time taken by them to cover the same direction is , 1/X : 1/Y i.e. Y:X
- The distance covered by a train in passing a pole , or a standing man , or a single post ,or any object (of some negligible length ) is equal to the length of the train.
- If a train of length l1 m passes through a bridge , a platform of length l2 m, the running train travels a distance (l1 + l2) m.
- If a train having a length of l1 m , and speed v1 m/s. Another train of length l2 m and speed v2 m/s are running on two parallel tracks in the same direction , then the faster train will over take slower in (l1 + l2) / (v1 - v2) s , if (v1 > v2 ) and in (l1 + l2 ) / (v2 - v1) s , if (v2 > v1)
- If two trains that are of length l1 and l2 km , moving in opposite direction with speeds u km/h and v km/h, respectively , then the time taken to cross each other , = (l1 + l2) / (u + v)

#### FORMULAE REGARDING TO BOATS AND STREAM -

#### -> IF THE SPEED OF A BOAT (OR A BODY ) IN STILL WATER BE X KM/H AND THAT OF STREAM BE Y KM/H, THEN

SPEED OF BOAT DOWNSTREAM = (X + Y) KM/H

SPEED OF BOAT UPSTREAM = ( X - Y) KM/H

**-> LET THE SPEED OF BOAT IN DOWNSTREAM = U KM/H , SPEED OF UPSTREAM =V KM/H ,THAN**

**(A)**SPEED OF BOAT IN STILL WATER =(U + V) / 2 KM/H

(B) SPEED OF STREAM (CURRENT) = (U - V ) / 2 KM/H

**Lets do some exercises now,**

SPEED, TIME AND DISTANCE EXERCISE 1 |

SPEED, TIME AND DISTANCE EXERCISE 2 |

SPEED, TIME AND DISTANCE EXERCISE 3 |

SPEED, TIME AND DISTANCE EXERCISE 4 AND HINTS AND SOLUTIONS 1 |

HINTS AND SOLUTIONS 2 |

HINTS AND SOLUTIONS 3 |

HINTS AND SOLUTIONS 4 |

HINTS AND SOLUTIONS 5 |

HINTS AND SOLUTIONS 6 |

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