Time, Speed and Distance

Distance = Speed$$\times$$Time

Speed = $$\frac{Distance}{Time}$$

Time = $$\frac{Distance}{Speed}$$

While covering the Speed in m/s to km/hr. multiply it by 3.6. It is because 1m/s = 3.6 km/hr

If the ratio of the speeds of A and B is a : b, then

• The ratio of the times taken to cover the same distance is 1/a : 1/b or b : a.
• The ratio of distance travelled in equal time intervals is a : b

                                       Average speed= $$\frac{Total Distance Travelled}{Total Time Taken}$$

If a part of a journey is travelled at speed $$S_{1}$$ km/hr in $$T_{1}$$ hours and the remaining part at speed $$S_{2}$$ km/hr in $$T_{2}$$ hours then.
                                       Total distance travelled= $$S_{1}T_{1}$$+$$S_{2}T_{2}$$ km
                                                      Average speed=$$\frac{S_{1}T_{1}+S_{2}T_{2}}{T_{1}+T_{2}}$$ km/hr

If $$D_{1}$$ km is travelled at speed of $$S_{1}$$ km/hr, and $$D_{2}$$ km is travelled at speed of $$S_{2}$$ km/hr then

                                                       Average Speed= $$\frac{D_{1}+D_{2}}{\frac{D_{1}}{S_{1}}+\frac{D_{2}}{S_{2}}}$$ km/hr

• In a journey travelled at different speeds, if the distance covered in each stage is constant, the average speed is the harmonic mean of the different speeds.
• Suppose a man covers a certain distance st x km/hr and an equal distance at y km/hr

Then the average speed during the whole journey is $$\frac{2xy}{x+y}$$ km/hr

• In a journey travelled with different speeds, if the time travelled in each stage is constant, the average speed is the harmonic mean of the different speeds.
• If a man travelled for a certain time at the speed of x km/hr and travelled for an equal amount of time at the speed of y km/hr then

Then the average speed during the whole journey is $$\frac{x+y}{2}$$ km/hr

If two objects( or persons) A and B are moving with speeds $$s_a\ \&\ s_b$$ km/h then the relative velocity between them 

A) when they are moving in the same direction is $$\left(s_a\ -\ s_b\right)$$ km/hr

B) when they are moving in the opposite direction is $$s_a\ +\ s_b$$ km/hr

CLOCKS:

• In a well functioning clock, both hands meet after every 720 / 11 Mins.
• It is because the relative speed of the minute hand with respect to the hour hand = 11/2 degrees per minute.

BOATS & STREAMS

• If the speed of water is 'W' and speed of a boat in still water is 'B'

  1. Speed of the boat downstream is B+W
  2. Speed of the boat upstream is B-W

The direction along the stream is called downstream. And, the direction against the stream is called upstream.

If the speed of the boat downstream is x km/hr and the speed of the boat upstream is y km/hr, then

                                  Speed of boat in still water= $$\dfrac{x+y}{2}$$ km/hr
                                  Speed of stream= $$\dfrac{x-y}{2}$$ km/hr

If the speed a person is x m/s, and the speed of the escalator is y m/s, then the relative speed is (x+y) m/s (if in same direction).

If they are moving opposite direction, then the relative speed is (x-y) m/s.

Circular Tracks

If two people are running on a circular track with speeds in the ratio a:b where a and b are co-prime, then

• They will meet at a+b distinct points if they are running in the opposite directions.
• They will meet at |a-b| distinct points if they are running in the same direction.
  

If two people are running on a circular track having perimeter I, with speeds m and n,

• The time for their first meeting = $$\frac{I}{(m+n)}$$

(when they are running in opposite directions)

• The time for their first meeting = $$\frac{I}{(|m-n|)}$$

(when they are running in the same direction)

If a person P starts from A and heads towards B and another person Q starts from B and heads towards A and they meet after a time 't' then,

                                                        t = $$\sqrt{x*y}$$

where x = time taken (after the meeting) by P to reach B and y = time taken (after the meeting) by Q to reach A.

A and B started st a time towards each other. After crossing each other, they took $$T_{1}$$ hrs, $$T_{2}$$ hrs respectively to reach their destinations. If they travel at constant speeds $$S_{1}$$ and $$S_{2}$$ respectively all over the journey, Then
                                                     $$\frac{S_{1}}{S_{2}}$$=$$\sqrt{\frac{T_{2}}{T_{1}}}$$

Linear Tracks(Races)

If A and B both participate in an L metre race, and A beats B by x metres(say), then the time taken by A to complete L metres is the same as the time taken by B to complete (L-x) metres.

In that case, if speed of A is a m/s and B is b m/s then $$\dfrac{L}{a}=\dfrac{L-x}{b}$$ 

So, the ratio of their speeds is $$\dfrac{a}{b}=\dfrac{L}{L-x}$$

TRAINS:

Two trains of length $$L_1$$ and $$L_2$$ travelling at speeds of $$S_1$$ and $$S_2$$ cross each other in

• $$\frac{L_1+L_2}{S_1+S_2}$$ if they are going in opposite directions

• $$\frac{L_1+L_2}{S_1-S_2}$$ if they are going in the same direction

• While converting the speed in m/s to km/hr, multiply it by 3.6. It is because 1 m/s = 3.6 km/hr

If the speed is increased by $$\frac{a}{n}$$, the time will reduce by $$\frac{a}{n+a}$$  and vice versa.

Or If the speed is decreased by $$\frac{a}{n}$$, the time will increase by $$\frac{a}{n-a}$$ and vice versa

Because the Product of speed and time is distance, which remains constant. Hence, if one of the speeds or times increases, the other has to decrease.

Circular Tracks

If two people are running on a circular track with speeds in the ratio a:b where a and b are co-prime, then

• They will meet at a+b distinct points if they are running in the opposite directions.
• They will meet at |a-b| distinct points if they are running in the same direction.
  

If two people are running on a circular track having perimeter I, with speeds m and n,

• The time for their first meeting = $$\frac{I}{(m+n)}$$

(when they are running in opposite directions)

• The time for their first meeting = $$\frac{I}{(|m-n|)}$$

(when they are running in the same direction)

When a person is on an escalator, he can move in either one of the two ways. 

If they are moving with the escalator, there  will be fewer steps to climb than the total (Same direction= less steps)
If they are moving against the escalator, there will be more steps to climb than the total (Opposite directions = more steps) 

Let us suppose that an escalator has a total of L stairs , and it churns out E stairs/second(moves at this speed). We consider a person  who can climb P stairs/second. 

When the person is moving on the escalator, he will move certain steps on his own, while the escalator moves. 

Let us assume he climbs D stairs on his own. The time taken to do this would be $$\frac{D}{P}$$ seconds. In this time, the escalator will churn out $$\frac{D}{P}\times E$$ stairs.

So, the total number of steps when the person is 

(A) Moving With the Escalator = steps climbed oneself + steps produced by escalator OR $$L=D+\frac{D}{P}\times E$$
(B) Moving Against the Escalator = steps climbed oneself – steps produced by escalator OR $$L=D-\frac{D}{P}\times E$$