# Time and Distance Online Test 2019, Time, Distance, Speed Concept

**Time and Distance Online Test:** Important Concepts & Short Tricks on Speed, Distance & Time with formula. Today we are going to discuss a very important topic Time, Speed and Distance. This concept is used extensively for questions related to different areas of CAT, GMAT and Bank exams. For example boats and streams, trains, clocks etc.

**Time and Distance Online Test**

Test | Questions | Launch Test |

Time and Distance Online Test Series 1 | 50 | Go to Test |

Time and Distance Online Test in Hindi | 30 | Go to Test |

Time and Distance Online Test Series 3 |

**Time, Speed and Distance: Concept**

We are providing you Important Short Tricks on Speed, Distance & Time which are usually asked in SSC Exams. Use these below given short cuts to solve questions within minimum time. These shortcuts will be very helpful for your upcoming SSC CHSL Exam.

**Important formulae and facts of Time and Distance**

**Speed** is a very basic concept in motion which is all about how fast or slow any object moves. We define speed as distance divided by time.

Distance is directly proportional to Velocity when time is constant.

- Speed Distance Time formula mathematically written as:-
**Speed**= distance/time

**Formula of Time :-time =** distance/ Speed

So Formula of time is, time is equal to distance upon speed.

**Formula of Distance:- Distance**= (Speed * Time)

**Distance** = Rate x Time

- To find rate, divide through on both sides by
*time:*

**Rate = Distance/Time**

(given in units such as miles, feet, kilometers, meters, etc.) divided by time (hours, minutes, seconds, etc.). Rate can always be written as a fraction that has distance units in the numerator and time units in the denominator, e.g., 25 miles/1 hour.*Rate*is distance

So distance is simply speed into time.

**Note:** All three formulae that formula of speed, formula of time and formula of distance are interrelated.

**Convert from kph (km/h) to mps(m/sec)**

For converting kph(kilometre per hour) to mps(meter per second) we use following formula

** x km/hr**=(

*x*∗5/18)

*m*/*sec***Convert from mps(m/sec) to kph(km/h)**

For converting mps**(meter per second)**to kph**(kilometre per hour)**we use following formula

*x m*

**/**= X *(18/5)

*sec*

*km*/*h*- If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by then to cover the same distance is :
**1/a : 1/b or b : a** - Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then,

the average speed during the whole journey is :-**2xy/(x + y)**

- Relation between time, distance and speed: Speed is distance covered by a moving object in unit time:
**Speed=****Distance covered/ Time Taken**

**Rule :**** 1:** Ratio of the varying components when other is constant: Consider 2 objects *A* and *B* having speed Sa, Sb.

Let the distance travelled by them are Da and Db respectively and time taken to cover these distances be Ta and Tb respectively.

**Let’s see** the relation between time, distance and speed when one of them is kept constant

- When speed is constant distance covered by the object is directly proportional to the time taken.

ie;**If Sa=Sb**then**Da/Db = Ta/Tb** - When time is constant speed is directly proportional to the distance travelled. ie;
**If Ta=Tb**then**Sa/Sb=Da/Db** - When distance is constant speed is inversely proportional to the time taken ie if speed increases then time taken to cover the distance decreases. ie; If
**Da=Db**then**Sa/Sb= Tb/Ta**

**Rule 2: We know that when distance travelled is constant, speed of the object is inversely proportional to time taken**

- If the speeds given are in Harmonic progression or HP then the corresponding time taken will be in Arithmetic progression or AP
- If the speeds given are in AP then the corresponding time taken is in HP

**Distance Constant**

- If distance travelled for each part of the journey, ie d1=d2=d3=…=dn=d, then average speed of the object is Harmonic Mean of speeds.

Let each distance be covered with speeds s1,s2,…sn in t1,t2,…tn times respectively.

Then t1 =d/s1

t2 = d/s2

tn =d/sn

**then, Average Speed**= [(d + d + d+ … ntimes)]/ [d/s1 + d/s2+ d/s3+ … d/sn

**Average Speed**= (n)/[(1/s1 + 1/s2+ …. 1/sn)]

**Time Constant**

- If time taken to travel each part of the journey, ie t1=t2=t3=…tn=t, then average speed of the object is Arithmetic

Let distance of parts of the journey be d1,d2,d3,…dn and let them be covered with speed s1,s2,s3,…sn respectively.

Then d1=s1 t , d2=s2t, d3=s3t, … dn=snt

**then , Average Speed=** [(s1/t+ s2/t+ …. sn/t)/(t + t+ … ntimes)]

**Average Speed=( s1+ s2+s3+ … + sn)/n**

**Relative Speed**

- If two objects are moving in same direction with speeds
*a*and*b*then their relative speed is**|a-b|** - If two objects are moving is opposite direction with speeds
*a*and*b*then their relative speed is**(a+b)**

**Time and Distance Problems**

**Problem 1**: A man covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?

**Solution**: Speed =Distance / Time

⇒ Distance covered = 600m, Time taken = 2min 30sec = 150sec

Therefore, Speed= 600 / 150 = 4 m/sec

⇒ 4m/sec = (4*18/5) km/hr = 14.4 km/ hr.

**Problem 2**: A boy travelling from his home to school at 25 km/hr and came back at 4 km/hr. If whole journey took 5 hours 48 min. Find the distance of home and school.

**Solution**: In this question, distance for both speed is constant.

⇒ Average speed = (2xy/ x+y) km/hr, where x and y are speeds

⇒ Average speed = (2*25*4)/ 25+4 =200/29 km/hr

Time = 5hours 48min= 29/5 hours

Now, Distance travelled = Average speed * Time

⇒ Distance Travelled = (200/29)*(29/5) = 40 km

Therefore distance of school from home = 40/2 = 20km.