quant-time-speed-distance-theory-part1

Time Speed and Distance

Your ability to solve the problems on relative motion, circular motion, and problems

based on trains, boats, clock, races etc. will depend only on the depth of your

understanding

1. Speed Distance/Time

2. Distance = Speed x Time

3. Time = Distance/ Speed

4. x km/hr = (x x 5/18) m/sec

5. y m/sec = (y x 18/5) km/hr

6. If a certain distance is covered at x km/hr and the same distance is covered at

y km/hr, then the average speed during the whole journey = km/hr

7. If a person changes his speed in the ratio m: n. then the ratio of time taken

becomes n : m

8. For the same distance speed varies inversely as time and in the same time

distance varies directly as speed

9. Average Speed =

Please note that the average speed of a moving body is NOT EQUAL to the average

of the speed

10. If a body covered one part of the journey at speed p and the remaining part of

the journey at speed q and the distances of the two parts of the journey are in the

ratio m : n, then the average speed for the entire journey is

Conversions

• To convert cm to metre divide by 100.
• To convert km to min. multiply by 1000.
• To convert hours to min. multiply by 60.
• To convert hours to sec multiply by 3600.
• To convert minutes to hours divide by 60.
• To convert seconds to hours divide by 3600.

Example 1.Time taken at 48 km/h is 24 minutes. What is the time if speed

decreases by 25%?

Solution:Speed ratio = 4: 3

Time ratio 3: 4 = 24: 32

Time ratio = 32 minutes

Example 2. Walking at 48% less speed, time taken is 75 minutes. Find the time

taken a usual speed.

Solution:Speed ratio = 100 : 52 25 : 13

Time ratio = 13 : 25 (speed and time vary inversely)

= 39 : 75

Time = 39 minutes

Example 3.A 240m long train travelling at 36 km/h clears a 600 m long platform in

how much time.

Solution: Speed of Train = 36 x 5/18 m/s 10m/sec.

Time = Distance/Speed 840m/10m/s = 84sec.

Example 4. A train at 148 km/h .overtakes a motorcyclist at 58 km/h in the same direction in 10 sec. What is the length of the train?

Solution:Distance = Speed x time

= (148 - 58) x 5/18 x 10m 250m

Example 5.Speed from A to B = 10 km/h and from B to A is 40 km/h. What is the average speed?

Solution:Average Speed = 2 x 10 x = 16 km/hr.

(If the distance is same then average speed is the harmonic mean)

Trains

Relative Speed:

The speed of one (moving) body in relation to another moving body is called the

relative speed of these two bodies, i.e., it the speed of one moving body as

observed, from the second moving body.

If two bodies are moving in the same direction, the relative speed is equal to the

difference of the speed of the two bodies.

If two bodies are moving in the opposite direction, the relative speed is equal to the

sum of the speeds of the two bodies.

• When a train crosses a tree/pole etc. Time taken to cross in seconds x speed in m/s = length of the train in m.

• When a train crosses a platform/tunnel/bridge etc. Time taken to cross in seconds x speed in m/s = length of the train in m + length of platform/tunnel/bridges in m.
• In problems on two trains the lengths of the trains are always added irrespective of whether the trains are running in same or opposite directions on parallel tracks.

• If trains are running in opposite directions.

         Time taken to completely cross each other =

• If trains are running in same directions

         Time taken by faster train to completely overtake the slower train

              =

• If two objects pass each other while travelling in the same direction then the

                time taken to pass =

Example 1.What is the time taken by a train running at 72 km/hr to cross a man

standing on a platform, the length of the train being 180 m?

Solution: Speed of train = 72 kmph = 72 x = 20 m/s

Distance = length of the train = 180 m

Time = = 9 sec

Example 2. How long will a train 100m long and traveling at a speed of 45 kmph,

take to cross a platform of length 250 m?

Solution: Distance = length of the train + length of the platform = 100 + 250 = 350m

Speed of the train = 45 kmph

= 45 x = 12.5 m/sec.

Therefore, Time = = 28 sec.

Example 3. Find the length of bridge, which a train 120 m long traveling at 54 kmph

can cross in 30 seconds.

Solution: Speed of the train = 54 kmph = 54 x = 15 m/sec

Distance covered in 30 seconds = 15 x 30 = 450 m

Length of bridge = Distance covered — Length of the train = 450 - 120 = 330 m.

Boat and Stream

Problems related to boats and streams are different in the computation of relative

speed from those of train/cars.

When a boat is moving in the same direction as the stream or water current, the

boat is said to be moving

WITH THE STREAM OR CURRENT.

When a boat is moving in a direction opposite to that of the stream or water current,

it is said to be moving

AGAINST THE STREAM OR CURRENT.

If the boat is moving with a certain speed in water that is not moving, the speed of

the boat is then be called the

SPEED OF THE BOAT IN STILL WATER.

When the boat is moving upstream, the speed of the water opposes (and hence

reduces) the speed of the boat.

When the boat is moving downstream, the speed of the water add (and thus adds to)

the speed of the boat.

Thus have

Speed of the boat against stream = Speed of the boat in still water - Speed of

the stream

Speed of the boat with the stream = Speed of the boat in still water + Speed of

the stream

Relative Speed

If u is the speed of the boat down the stream and v is the speed of the boat up the

stream, then we have the following two relationships,

• Speed of the boat in still water =

• Speed of the water current =

Following relations shall also be useful

• Speed of boat in still water = [speed downstream + speed upstream]

• Speed of stream = [speed downstream - speed upstream]

• If speed of boat in still water is b km/hr, speed of river or stream is s km/hr, speed downstream is d km/hr and speed upstream is u km/hr then

                d = (b + s) km/hr and u = (b - s) km/hr

                   b = (u + d) km/hr and s = (u - d) km/hr

NOTE : In problems, instead of a boat, it may be a swimmer but the approach is

exactly the same, instead of boats/swimmers in water. It could also be a cyclist

cycling against or along the wind. The approach to solve problems still remains the

same.

Example 1. A boat travels 30km upstream in 6 hours and 20 km downstream in 5

hours. Find the speed of the boat in still water and the speed of water current.

Solution: Upstream speed = 30/6= 5 kmph

Downstream speed = 20/5 = 4 kmph

Speed in still water = (5 + 4)/2 = 4.5 kmph

Speed of the water current = (5 - 4)/2 = 0.5 kmph

Example 2. A man can row 8 km in one hour in still water. If the speed of the water

current is 2 km/hr and it takes 3 hours for him to go to a new place and return, find

distance from the starting point to the new place.

Solution: Let the distance be x km

Upstream speed = 8 - 2 = 6 kmph

Downstream speed = 8 + 2 = 10 kmph

Total time = time taken upstream + time taken down stream = kmph

x = 11.25 km

RACES & TRACKS

When two persons P and Q are running a race, they can start the race at the same

time or one of them may start a little later than the other. In the second case

suppose P starts the race and after 5 seconds, Q starts. Then we say P has a “start”

of 5 seconds. Alternatively, in a race between P and Q. P starts first and then when

P has covered a distance of 10 meters, Q starts. Then we say that P has a “start” of

10 meters.

In a race between P and Q where Q is the winner, by the time Q reaches the winning

post first. If P still has another 15 meters to reach the winning post, then we say that

Q has won the race by 15 meters. Similarly, if P reaches the winning post 10

seconds after Q reaches it, then we say that Q has won the race by 10 seconds.

In problems on RACES, we normally consider a 100 m race or a 1 km race. The

length of the track NEED NOT necessarily be one of the two figures mentioned

above but can be as given in the problem.

When two or more persons running around a circular track (starting at the same

point and at the same time), then we will be interested in two main issues:

• When they will meet for the first time and

• When they will meet for the first time at the starting point

To solve the problems on circular tracks, you should keep the following points in

mind.

When two persons are running around a circular track in OPPOSITE directions

• The relative speed equal to the sum of the speeds of the two individuals and

• From one meeting point to the next meeting point, the two of them TOGETHER cover a relative distance equal to the length of the track.

When two persons are running a circular track in the SAME direction.

• The relative speed is equal to the difference of the speeds of the two individuals and

• From one meeting point to the next meeting point, the faster person covers one COMPLETE ROUND more than the slower person.

Please note that when we have to find out the time taken by the persons to meet for

the first time at the starting point, what we have to do is to find out the time taken by

each of them to complete one full round and then take the LCM of these two timings

(L/a and L/b are the timings taken by the two of them respectively to complete on

full round).

When three people are running around a circular track

Let the three people A, B and C with respective speeds of a, b and c (a > b > c) be

running around a circular track (of length L) starting at the same point at the same

time In this case we consider the three persons running in the same direction as the

general case.

Example 1: In a race of 1000 m. A beats B by 50 m or 5 seconds.

Find (i) B’s speed (ii) A’s speed (iii) Time taken by A to complete the race.

Solution: Since A beats B by 50 m, it means by the time A reaches the winning

point, B is 50 m away and as A beats B by 5 seconds, it means B takes 5 seconds to

reach the winning point. This means B covers 50 m in 5 seconds i.e., B’s speed is

50/5 = 10 m/s. Since A wins by 50 m, in the times A covers 1000 m, B covers 950 m

at 10 m/s, B can cover 950 m in 950/10.

i.e., 95 seconds i.e., 1 min 35 seconds.

A completes the race in 1 min 35 seconds

A’s speed is = m/s

Example 2. Rakesh runs times as fast as Mukesh. In a race, if Rakesh gives a

lead of 60 m to Mukesh. Find the distance from the starting point where both of

them will meet.

Solution: Since Rakesh runs times as fast as Mukesh, in the time Mukesh runs

3 meters, for every 4 meters he runs.

Since he has given a lead of 60m, he will gain this distance by covering 4 x 60 =

240m

Hence they will meet at a point 240 m from the starting point

Example 21 In a 1500 m race, Tarun beats Manoj by 150 m and in the same race

Manoj beats Rahul by 75 m. By what distance does Tarun beat Rahul

Solution: Let us write the data given as below

This will be equal to 1350 x = 1282.5

But Manoj running 1350 m is the same as Tarun running 1500m. Hence, Tarun beats

Rahul by m.

Other Quants Notes:

• Number System
• Average
• Percentages
• Profit & Loss
• Simple and Compound Interest
• Ratio & Proportion
• Partnership
• Mixtures & Allegation
• Time, Speed & Distance
• Work, Pipes & Cistern
• Mensuration
• Geometry
• Height & Distance
• Progression
• Permutation & Combination
• Probability
• Data Interpretation