Shortcuts and Key concepts of Time ,Speed and Distance

 

             TIME,SPEED & DISTANCE – TSD

In this post, you will get complete information about the most important topic of time, speed, and distance. In this topic, we will see all concepts of time, speed, and distance and all types of questions that come in all competitive exams especially staff selection commission,banking (Ibps), Csat, and TCS campus placement drive exams.

TSD- We all are very familiar with all three terms time, speed, and distance  since our childhood but what is the application of all three terms from the exam perspective, we will see in this chapter.

 

Time -  It had never been  stopped and will not be stopped until the earth exists so, we can easily say it is only increasing function in this existing earth  But in this chapter, we will see sometimes, we will make time constant in the interval of two points which will very interesting so stay tuned till last of this chapter. 
 

Speed -  Speed, we know it is in one hour/in unit time, how much you Travelled is called speed of that person. It is expressed generally in kilometers per hour. It has often been seen in examinations speed will be given either in m/s (meter per second) or in kilometers per hour. So how we will change from kilometer per hour to meters per second and vice-versa.

CONVERSION FROM KILOMETER PER HOUR TO METER PER SECOND- 

Whenever you have to convert from km/hr to m/s, you need to multiply it by 5/18. Remember two numbers in your mind one is 5 and another is 18. Sometimes, while attempting competitive exams, we forgot which number will come in the numerator part and which no. will come in the denominator part. Look km/hr is a big unit than m/s .so when you need to convert from a bigger unit to a smaller one just keep a smaller no. (5) in the numerator part while the other (18) in a denominator part so, finally we multiply the bigger unit (km/hr) with 5/18 to get the smaller unit (m/s). We can do just reversed while converting smaller unit(m/s) to bigger unit (km/hr) by keeping the bigger number(18) in the numerator part and a smaller number(5) in the denominator part so, finally we multiply m/s with 18/5 to get speed in km/hr. We can best understand this with the figure below.    

Example1. If the speed of a person is 36 km/hr. What will be his speed in m/s? 

Sol- Just multiply with 5/18.

       36 × 5/18  = 10 m/s. 

Example 2. Convert 10 m/s , 40 m/s  and 60 m/s into Km/hr. 

Sol- 10 m/s = 10 × 18/5 = 36 km/hr

       40 m/s= 40 × 18/5  = 144 km/hr

        60 m/s = 60 × 18/5 = 216 km/hr 

DISTANCE - It is nothing but the whole path travelled without making any shortcut /displacement. Normally we measure distance in their SI Unit meter.

 The short distance we measure in meters.

 Long-distance we measure in kilometer.

 The very short distance we measure in cm generally

 Conversion chart – We can easily understand through the given figure, it will help you to solve questions quickly in your exam time.

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Relation among three terms - 

 The whole concept entirely depends upon only one basic relation which is 

                    DISTANCE =  SPEED × TIME  

Based on it, we can find the remaining two which is shown below -

                    SPEED = DIS/ TIME ----------------------------------------------1

                    TIME =  DIS /SPEED  ---------------------------------------------2

We will only keep one relation in mind which is   “dis = speed × time ". if anything apart from distance is being asked then just keep distance in numerator part and rest of what we left will keep  in the denominator part, you can easily see in equations 1 and 2.

SOME IMPORTANT KEY CONCEPTS – CASE 1-  WHEN  DISTANCE = Fixed  Value /Constant 

1.    WHAT WILL HAPPEN WHEN IN BETWEEN TWO POINTS  /TWO PLACES  - WHENEVER IF IN A QUESTIONS, IT HAS BEEN GIVEN THAT IN BETWEEN TWO POINTS OR IN BETWEEN TWO PLACES  THEN ALWAYS KEEP IN MIND THAT THERE IS CONSTANT /FIXED DISTANCE BETWEEN TWO POINTS. WHENEVER DISTANCE IS FIXED THEN 

IF DISTANCE = FIXED BETWEEN TWO PLACES/POINTS THEN

                                 I.             SPEED = 1/TIME     

                              II.             TIME = 1/SPEED

                              III.            SPEED IS INVERSELY PROPORTIONAL TO  TIME.

                             IV.            IN TERMS OF RATIO -    IF RATIO OF SPEED IN GIVEN TWO POINTS (SUPPOSE FROM PATNA TO DELHI)    IT IS  A: B THEN RATIO  OF THE  TIME  WILL  BE  B: A.

 

                                 V.            IN TERMS OF RATIO – IF RATIO OF TIME IS GIVEN IN FIXED INTERVAL /DISTANCE   C: D THEN RATIO OF SPEED WILL BE D: C.

CASE2 -  WHEN  SPEED IS FIXED IN THE WHOLE JOURNEY /PATH.

  

              DIS= SPEED × TIME
                       DIS = CONSTANT ×TIME

                       I.e.  Distance is directly proportional to the time means, if distance increase then time will increase and vice versa. 

Ø  If distance ratio is given in a question a: b then the same ratio will be for time because both are proportional to each - other.

Distance -    a:    b

  Time -        a    :  b 

Ø If time ratio is given in the above context suppose it is a: b then, speed ratio will be also a: b .

 

Time ratio -   a:  b

 

Speed ratio -  a :   b  

CASE 3 -  WHEN TIME IS KEPT CONSTANT / IN EQUAL INTERVAL FOR JOURNEY –

 

         DIS  =  SPEED ×  TIME  

   DIS =  SPEED × CONSTANT (VALUE)

  SO, DISTANCE IS DIRECTLY PROPORTIONAL TO  SPEED. 

Ø IF SPEED RATIO IS GIVEN  a: b then, distance ratio will be also a: b.

 

Speed -      a: b

Distance -  a: b

Ø Similarly,  if the distance ratio is given  a: b then  , speed ratio will be also  a: b 

AVERAGE SPEED -   IT IS THE RATIO OF  TOTAL DISTANCE COVERED IN TOTAL TIME.

 

AVG SPEED =   TOTAL DISTANCE IN JOURNEY / TOTAL TIME TAKEN DURING THE JOURNEY.

APPLICATION OF ALL ABOVE CONCEPTS WITH PROBLEMS 

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TYPE 1 – WHEN DISTANCE IS FIXED 

Q1. An airplane covers   a certain distance at a speed of 240 km/hr in 5 hour, To cover the same distance in 5/3 hours it must travel at a speed of

  a.       300 km/hr

         b.      360 km/hr

     c       c.       600 km/hr

 d.        d.  720 km/hr

 

Sol-     Here , easily we can find distance by reading first line statement of questions in each hour 240 km travels so , in 5 hour =  240 × 5 = 1200 km

So , distance = 1200 which is fixed  in another trip from the questions

 Time = 5/3 hours

So ,   speed =  1200 / (5/3)       =  1200 ×3/5  =  720 km/hr

 Second method –

Here distance is fixed in journey = 240 × 5 = 1200 km 

So , we can write 

Time -         5   :    5/3

           Or,      15  : 5 

           or,       3: 1

                 Since 1 unit/part = 240 km/hr so , 3 unit/part  = 3 ×240 = 720 kmphr

Note – Try to use second method that will save your valuable time  in exam and less stress.

Q2.  Walking at ¾ of his usual speed, a man is 3/2 hours late. His usual time to cover the same distance in hours is?

a.       9/2 hours

b.      11/2 hours

c.       4 hours

d.      5 hours 

Sol-  

 Here , again distance in a  given interval is fixed  so go through ratio method to solve easily  

Here , time difference in terms of ratio =  4-3 = 1 unit /part

 And in questions it is equal to =  3/2 hours

       So , 1 unit = 3/2 hours 

              So, 4 unit =  3 × 3/2 =  9/2 =  4.5  hours  

 

Note - Always take denominator value here it is 4 part as usual speed part   according to sentence so, numerator part will become normal  speed part  which is  3 part. This approach will give you less time and faster solution which is most important in competitive exams.