Time, Speed and Distance - Atreya Roy
Content & PR team - MBAtious
Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.
Speed of a particle / body is the Distance it covers per unit time.
Hence Speed = Distance / Time
Average Speed :
If a body traverses a distance at different speeds, then the average speed is calculated as:
Total Distance Covered / Total Time taken
Mathematically Average Speed = D / [( D1/S1) +(D2/S2) +(D3/S3)…..]
Where D1,D2,D3,D4… S1,S2,S3,S4… are the distances and their respective speeds
D= D1+D2+D3+D4….. is the total distance
A car travels 800 km at a speed of 50 kmph , another 600km at a speed of 30 kmph . What is the average speed of the car in the whole journey ?
Average Speed = Total Distance / Total Time
= (600+800)/ ((800/50) + (600/30) ) = 1400 / 16 +20 = 1400/ 36 = 38.89 Kmph
Remember : The distance always remains same.
If a body with speed S1 takes time T1 and another body with a speed S2 and time T2 cover the same distance then, S1*T1 = S2*T2 = D
Relative Velocity :
The concept of relative velocity is dependent on two bodies in movement. Movement can be both in the same direction or in opposite directions.
Let body A have a speed of V1 and body B have a speed of V2.
If the two bodies start from the same point in the same direction, then their relative speed = V1~ V2
Where ~ signifies the difference and hence the difference in the speed of the two body.
Similarly, if the start from the same point and travel in opposite directions, then their relative speed will be (V1+V2). Means after every unit time, they will be (V1+V2) distance units away from each other.
Relative Speed :
When two bodies travel in the same direction = | V1 – V2 |
When two bodies travel in opposite directions = V1+ V2
Runners A and B start from two end points of a road XY with speeds 10 kmph and 15 kmph and run towards each other. The length of the road is 100 Kms. After how many hours from the start will they meet. After what individual distances travelled by both will they meet ?
Total Distance of the track XY = 100 km
Speed of A = V1 = 10 kmph
Speed of B = V2 = 15 kmph
They run in opposite direction hence Relative speed = 10+15 = 25 kmph
Time taken to meet = Distance of the track / Relative Speed = 100/25 = 4 hours
Hence they will meet after 4 hours.
Distance covered by A = Speed of A * time = 10*4 = 40 km
Distance covered by B = Speed of B * time = 15*4 = 60 km
Note : The distance covered by both is always equal to the total distance of the scenario taken under consideration.
Another type of question that one might face in the exams is Train related problems. Now we will learn the various types of train related sums that might appear in the exam. We will face problems where the train passes an object or a person or another train, in that scenario, passes means the tail of the train passes the tail of the body.
Distance covered by the train = Length of train + Length of the object.
We will again take into consideration the theory of relative speed as two bodies are moving here. Hence if they move in the same direction, Relative Speed = V1-V2 , and if they move in the opposite direction, Relative speed = V1+V2, where V1 and V2 are the speeds of the train and the other body respectively.
For stationary objects, Relative speed = Speed of the train.
Remember : When two trains are moving in the same direction, from the same point, the slower train cannot overtake the faster one.
A train crosses a man standing on a platform in 20 seconds. The Speed of the train is 72 Kmph. What is the length of the platform if it crosses the platform in 50 seconds ?
Speed of the train = 72 kmph = 72 *5/18 = 20 m / s
Time taken to cross the man = 20 seconds.
Hence the length of the train = 20*20 = 400 m
Now the train will cross the platform in 50 seconds. According to the formula,
(Length of train + Length of platform) / Speed of train = Time Taken
( 400 + x ) / 20 = 50
X= 1000 -400 = 600 m
Hence the length of the platform is 600 metres.
Stoppage Time For a Train :
If the speed of the train is V1 kmph without stops and V2 with stops, then
Stoppage time per hour = ( V1 - V2 ) / V1 hour
A train travels at a speed of 100 kmph without stopping in any of the intermediate stations between two places. When it stops in few of the intermediate stations, the average gets reduced to 80 kmph. For how many minutes does the train stop per hour at an average ?
Normal average speed of train = 100 kmph
Reduced average speed of train = 80 kmph
Stopping time per hour = (100-80) / 100 = 20/100 = 1/5 hour = 12 minutes
Between 2 stations A and B, there run two trains X and Y from A to B and from B to A and again back. One day, they both started from A at the same time. Speed of X is greater than Speed of Y. It is also known that the ratio of their speeds is 2:5. If the distance between A and B is 350 kms. When will Y overtake X after the start of the journey ?
Speed of X > Speed of Y
Since both leave A at the same time, Y wont be able to overtake X.
Boats and Streams
While doing the sums on Boats and streams we will come across various terms like : speed of river, speed of boat, upstream, downstream. Lets see what each of them mean :
- Speed of stream/river : it is the speed with which the current flows.
- Speed of boat : the speed of the boat is considered to be its original moving speed when it is placed in still water without no external support of movement.
- Upstream : Upstream means going from to the top of the stream from the bottom of the stream.
- Downstream : it means going from top to the bottom of the specified distance in the stream.
So it is evident from the above points that when speeds are considered,
- In upstream movement since the boat is going from bottom to top and river flows from top to bottom, the speed of the boat will decrease due to the opposite flow of the river. Hence New Speed = Speed of boat – Speed of River
- In downstream movement since the boat is going with the current, from top to bottom, then their speed will be added and movement will be faster. New Speed of boat = Speed of Boat + Speed of River
So, Mathematically , we will look like this :
If the speed of man rowing in still water is = x
Speed of stream/river = y
Speed of Boat in Downstream = x+y
Speed of boat in Upstream = x-y
Always remember the speed of rowing in still water must be greater than speed of river, or Upstream movement is not possible
So, formulating the above 2 ideas, Mans rowing speed in still water = 1/2 * (2x) = 1/2 * ( (x+y) +(x-y) ) = 1/2 * ( Downstream speed + Upstream Speed)
Speed of stream = y = 1/2 * 2y = 1/2 ( (x+y) – (x-y) ) = 1/2 * (Downstream – Upstream)
Here, downstream and upstream refers to their rates.
Important Shortcut :
When Distance in downstream and upstream are same,
Total Time = (Speed in still water * Total Distance) / ( Upstream rate * Downstream Rate )
Total Distance = 2 * one side distance.
A person can swim down the river at 8km/hr and up the river at 2 km/hr. What is the speed of the man in still water. Also find the speed of stream .
Man’s rate in still water = 1/2 ( Upstream + downstream ) = 1/2 (8+2) = 5 km/hr
Speed of river = 1/2 ( Downstream – Upstream) = 1/2 (8-2) = 3km/hr
A man can row 4.5 km/hr in still water. It takes him twice as long to row upstream as to row downstream. What is the rate of the current?
Speed of boat in still water (b) = 4.5 km/hr. It is given upstream time is twice to that of down stream.
Downstream speed is twice to that of upstream. So b + u = 2(b – u) So, u=b/3 = 4.5/3
= 1.5 km/hr.
A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the stream.
If a is rate downstream and b is rate upstream
Rate in still water = 1/2(a+b)
Rate of current = 1/2(a-b)
= > Rate in still water = 1/2(20+10) = 15 kmph
= > Rate of current = 1/2(20-10) = 5 kmph
A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is
Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr
So we know from question that it took 4(1/2)hrs to travel back to same point.
So, 30/(15+x) – 30/(15−x)=4 (1/2)
= 900/ (225−x^2)=9/2
= > 9x^2=225= > x=5km/hr
A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is
Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18 )*(18/5) kmph
= 5 kmph