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Speed,Time and Distance (STD)

Basic Formulas

Speed = Distance/Time
Time = Distance/speed
Distance = speed*time

Conversion From km/hr to m/s and m/s to km/hr

To convert Kilometers per Hour(km/hr) to Meters per Second(m/s) x km/hr=x×5/18m/s
To convert Meters per Second(m/s) to Kilometers per Hour(km/hr) x m/s=x×18/5 km/hr

Important Relations:
Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝ 1/Time (when distance is constant)
If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a
suppose a man covers a distance at x kmph and an equal distance at y kmph.then the average speed during the whole journey is (2xy/x+y) kmph


1.  If two objects are moving in opposite directions towards each other  at speeds u and v, then
 relative speed = Speed of first + Speed of second = u + v.
This is also the speed at which they are moving towards each other or the speed at which they may be moving away from each other.

2. If the two objects move in the same direction with speeds u and v, then
relative speed = difference of their speeds = u – v.
This is also the speed at which the faster object is either drawing closer to the slower object or moving away from the slower object as the case may be.

3. If the two objects start from A and B with speeds u and v respectively, and after crossing each other take a and b hours to reach B and A respectively, then u : v = √b/a

Important Points:

1. 1 Pole and I Train:
Length of The Train (m) = Speed of the Train (m/s) × Time taken to cross the pole (s)

2. 1 Train and 1 Platform:
Length of the Train + Length of the Platform (m) = Speed of the Train (m/s) × Time taken to cross the platform (s)

3. 1 Train with speed speed v1  and 1 moving person with speed v2
Case 1: If both are moving in same direction
Length of The Train (m) = [Speed of the Train - Speed of the Man] (m/s) × Time taken to cross the man (s)

Case 2: If both are moving in opposite direction
Length of The Train (m) = [Speed of the Train + Speed of the Man] (m/s) × Time taken to cross the man (s)

4. 2 Trains with speeds v1,v2
Case 1: If both are moving in same direction

[Length of The Train 1 + Length of the Train 2](m) = [Speed of the Train1 - Speed of the Train 2] (m/s) × Time taken to cross (s)

Case 2: If both are moving in opposite direction
[Length of The Train 1 + Length of the Train 2](m) = [Speed of the Train1 + Speed of the Train 2] (m/s) × Time taken to cross (s)

Type 1. Basic Concept
Car covers a certain distance at 36 km/hr. How much distance does he cover in 3 min?
Speed = 36 km/hr = 36 × 5/18 × m/s = 10 m/s
Thus, the distance covered in 3 min = (10 x 3 x 60) = 1,800 m.

Type2:  with n/y Usual Speed
Walking at 3/4 of his usual speed a man is 1 1/2 hr late. Find his usual travel time. ( SSC CGL – 2013)
Let usual time be t hours. We know that S X T = D, But now his speed is 3/4 (S).  So time to cover the same distance becomes 4/3 (T).
4/3 × t = t + 3/2
t = 4.5 hr.

Second method:
We know that when his speed got reduced his time went up by 1 1/2 hr.
st =  3s/4(t +3/2 )
⇒ 4st = 3st + 4.5 s
⇒ t = 4.5 hr.

Type3: Same direction at Different Speed
A boy started from his house by bicycle at 10 a.m. at a speed of 12 km per hour. His elder brother started after 1 hr 15 mins by scooter along the same path and caught him at 1.30 p.m. The speed of the speed of the scooter will be (in km/hr) (SSC FCI Assistant Grade – III -2013  
(a)4.5                                    (b) 36                  
(c) 181/2                              (d) 9
Distance covered by scooter in 2 1/4 hours
12× 7/2=x×9/4
56/3=182/3 kmph

Type4.Two Train in Opposite Direction
Two trains, each of length 125 metre, are running in parallel tracks in opposite directions. One train is running at a speed 65 km/hour and they cross each other in 6 seconds. The speed of the other train is ( SSC 10 + 2- 2013)
Length of both trains = 250 metre
Speed of second train = x kmph = (65+x) kmph
= 65 +x) ×5/18 m/sec.
Time = (Sum of lengths of trains)/(Relative speed )
6 = 250/((65+x)×5/18)
6× 5/18×(65+x)=250
65+ x=(250×3)/5
65 + x=150
x=150-65=85 kmph

Type5.Same Distance with different times
Two cycle cover the same distance in 15 km/hr and 16 km/hr, respectively. Find the distance travelled by each, if one takes 16 min longer than the other does.
Let the required distance be x km.
x/15−x/16=16/60 or 16x – 15x = 64 or x = 64
Hence, the required distance = 64 km

Second Method:
If the speeds are in the ratio 15:16 then the ratio of the times to cover the same distance would be in the ratio 16 : 15 or 16x and 15x respectively.  But we know that difference in the times is 16 min or 16x - 15x = 16 min so x = 16 min
From the above derivation, second person takes 15 x 16 min = 4 hours to cover the distance at the speed of 16 km/hr.  So distance = 4 X 16 = 64 km

Type 6: Based on Difference in time / distance
A hare makes 9 leaps in the same time as a dog makes 5. But the dog’s leap is 2m while hare’s is only 1 m. How many leaps will the dog have to make before catching up with the hare if the hare has a head start of 16 m?
Distance covered by dog in 5 leaps = 5 × 2 = 10 m
Distance covered by hare in 9 leaps = 9 × 1 = 9 m
Distance gained by the dog in 5 leaps = 1 m. Hence, for 1 m gain he has to make 5 leaps.
Number of leaps required by the dog to gain 16 m = 5 × 16 = 80 leaps.

Type 7: Question Based on better than another train
Train A took 35 minute to cover a distance of 50 km. If the speed of train B is 25% faster than train A, it will cover the same distance in:
Let, speed of train A = 100 km/h
Then, speed of train B = 125 km/h
Ratio of speeds of trains A and B = 100 : 125 = 4 : 5
Therefore, Ratio of time taken by them to cover equal distance = 5 : 4
Given, time taken by train A = 35 minute
Time take by train B = 4/5 x 35 minute = 28 minute.

Second Method:
Speed of train B is faster by 25% = 1/4
Therefore, Decrease in time taken by train B = 1/5 x 35 minute = 7 minute.
Therefore, Time taken by train B = 35 - 7 = 28 minute.

Type 6:
A man is walking at a speed of 10 km/h. After every km, he takes rest for 4 minutes. How much time will he take to cover a distance of 10 km?
He covers 10 km in 1 hour (i.e. in 60 minutes)
Therefore He will take 6 minutes in covering 1 km.
He rests for 4 minutes after every km.
Time taken = (6 + 4) minutes = 10 minutes for every km.
Therefore, Time taken (for first 9 km) = 9 x 10 = 90 minutes.
Time taken to cover 10th km = 6 minutes
Therefore, Total time taken = 90 + 6 = 96 minute.
Hint: Rest time after 10th km is not added as he has reached his destination.

Type 8:
A train left station A for station B at a certain speed. After travelling for 100 km, the train meets with an accident and could travel at 4/5th of the original speed and reaches 45 minutes late at station B. Had the accident taken place 50 km further on, it would have reached 30 minutes late at station B. What is the distance between station A and B?
Let, initial speed of the train = 5 km/h
Then, speed after the accident = 4/5 x 5 = 4 km/h
Time taken to cover 50 km @ 5 km/h = 10 hours
Time taken to cover 50 km @ 4 km/h = 12 1/2 hours
Difference between times taken = 12 1/2 - 10 = 2 1/2 hours = 150 minutes
But, actual difference = (45 - 30) minutes = 15 minutes
= 1/10 of 150 minutes
Therefore, Speed is 10 times of assumed speed.
Therefore, Speed before accident = 10 x 5 = 50 km/h
And, Speed after accident = 10 x 4 = 40 km/h
Distance after accident is covered @ 40 km/h instead of 50 km/h
And, time difference = 45 minutes = 3/4 hour
Therefore, Distance between place of accident and B = (50×40)/(50−40)×3/4 = 150 km
Therefore, Distance between A and B = 100 + 150 = 250 km.

Type 9: Based On Average Speed
One third of a certain journey is covered at the rate of 25 km/hr one-fourth at the rate of 30 km/hr and the rest at 50 km/hr. The average speed for the whole journey is (FCI – 2012)
(a) 35 km/hr                       (b) 331/3 km/hr
(c) 30 km/hr                       (d) 37 1/(12 ) km/hr
(b) Let the total distance be x km.
Total time = (x/3)/25+(x/4)/30+(5x/12)/50
= x/75+x/120+x/120
= x/75+x/60=(4x+5x)/300=3x/100 hours
 Average speed
= (Total distance)/(Time taken)
= x/(3x/100)=100/3=33 1/3 kmph
Type 10:Based on Same direction
Two trains of equal length are running on parallel lines in the same direction at 46 km/h and 36 km/h. The faster train passes, the slower train in 36 seconds. The length of each train is : (SSC – 10+2- 2012)
Let the length of each train be x metre.
Relative speed
= 46 – 36 = 10 kmph
= (10×5)/18 metre/second
= 25/9 metre/second
x =50 metre


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