CAT Quant : Speed, Distance and Time – Important Formulas and Concepts
Speed
Distance covered per unit time is called Speed. $$Speed = \frac{Distance}{Time} = \frac{D}{T}$$
Proportionality Implicit in the Equation
Direct Proportionality between Time and Distance [When Speed is Constant] – In the formula (S * T = D), if Speed is constant, then $$T \propto D \ \Longrightarrow \ Ratio \ of \ \frac{D}{T} \ will \ be \ constant $$
Direct Proportionality between Speed and Distance [When Time is Constant] – In the formula (S * T = D), if Time is constant, then $$S \propto D \ \Longrightarrow \ Ratio \ of \ \frac{D}{S} \ will \ be \ constant $$
Inverse Professionality between Speed and Time [When Distance is Constant] – In the formula (S = D/T), if Distance is constant, then $$S \propto \frac{1}{T} \ \Longrightarrow \ Product \ of \ \ {S}*{T} \ will \ be \ constant $$
Conversion
To convert Speed in kmph to m/s — multiply by 5/18
To convert Speed in m/s to kmph — multiply by 18/5
Average Speed
Average Speed of a body travelling at different speeds is defined as – $$Average \ Speed = \frac{Total \ Distance \ Travelled}{Total \ Time \ Taken}$$ Note – Average Speed of a body is not always equal to the Average of their Speeds
If a body travels from point A to B with a speed of “p” and back to point A with a speed of “q” then, $$Average \ Speed = \frac{2pq}{p+q} $$ Note – The average speed does not depend on the distance between in this situation
If a body covers part of the journey at speed “p” and the remaining part of the journey at speed “q” and the distance of the two parts of journey are in the ratio “m : n”, then $$Average \ Speed = \frac{(m+n)pq}{mq + np} $$
In general, if a person travelling between two points reaches “p hours” late travelling at a speed of “u kmph” and reaches “q hours” early travelling at a speed of “v kmph”, the distance between them is given by $$Distance = \frac{uv}{v-u} * (p+q)$$
Acceleration
Acceleration is the rate of change of speed. It can be positive (+ve speed increase) or negative (-ve speed decrease). $$Final \ Speed = Initial \ Speed + (Acceleration * Time)$$
Relative Speed
The speed of one moving body in relation to another moving body is called relative speed of these two bodies
Same Direction – Two bodies A and B moving in the same direction with speeds V1 and V2 respectively, $$Relative \ Speed = | V_1 – V_2| \\ [A (V_1) \longrightarrow \ , \ B (V_2) \longrightarrow]$$
Opposite Direction – Two bodies A and B moving in opposite directions with speeds V1 and V2 respectively, $$Relative \ Speed = |V_1 + V_2| \\ [A (V_1) \longrightarrow \ , \ \longleftarrow B (V_2) ]$$
Problems on Trains
Trains are a special case of application of Time, Speed and Distance. Some common cases are discussed below.
CASE 1 – Train crossing a stationary object without length. $$Time = \frac{L_t}{V_t} = \frac{Length \ of \ Train}{Speed \ of \ Train}$$
CASE 2 – Train crossing a stationery object with length. $$Time = \frac{L_t + L_o}{V_t} = \frac{Length \ of \ Train + Length \ of \ Object}{Speed \ of \ Train} $$
CASE 3 – Train crossing a moving object without length $$Time = \frac{L_t}{V_t \pm V_o} = \frac{Length \ of \ Train}{ Relative Speed \ of \ Train \ and \ Object}$$
CASE 4 – Train crossing a moving object with length. $$Time = \frac{L_t + L_o}{V_t \pm V_o} = \frac{Length \ of \ Train + Length \ of \ Object}{Relative \ Speed \ of \ Train \ and \ Object} $$
Special Scenario – If two moving objects are approaching each other after crossing they take a and b seconds to reach their destination respectively, then $$Meeting \ Time \ (T) = \sqrt{ab}$$
Boats and Streams
Boats and Streams are another special case of application of Time, Speed and Distance.
Speed of Boat in Still Water (SB)
Speed of Stream (Ss)
Up Stream – When boat and stream travel in opposite directions (against each other) $$T_{UpStream} = \frac{D}{S_B -S_S}$$
Down Stream – When boat and stream travel in same direction (with flow of water) $$T_{DownStream} = \frac{D}{S_B + S_S}$$
If “u” is speed of boat in downstream and “v” is speed of boat is upstream then we have – $$Speed \ of \ Boat \ in \ Still \ Water \ [S_B] = \frac{u+v}{2} \\ Speed \ of \ Water \ Current \ [S_S] = \frac{u-v}{2}$$
Circular Motion
A special case of movement is when two or more bodies are moving around a circular track.
Opposite Directions – When two persons are running around a circular track in opposite directions, then
Relative Speed = [V1 + V2]
From one meeting point to the next meeting point, the two of them cover a distance equal to the “Length of track“ together.
Same Direction – When two persons are running around a circular track in the same direction, then
Relative Speed = [V1 – V2]
From one meeting point to the next meeting point, the faster person covers one complete round more than the slower person.
Note – The number of distinct meeting points while moving along the circular track –
Opposite Direction = [a + b]
Same Direction = [a – b]
where, “a : b” is the simplest ratio of their speeds.
When Two / Three people are moving around a Circular Track – Let two people A , B or three people A , B and C with respective speeds a, b and c (a>b>c) be running around a circular track of length “L” starting at the same point and at the same time –
Here, in 2nd Row, we have to find out the time taken by each of them to complete one full round and then take LCM of them,
Escalators
Time, Speed and Distance is applicable in certain cases of Escalators as well. There can be three different cases under escalators. they are as –
Escalator and Person move in Same Direction – [M + E]
Escalator is Switched off (Not Moving) – M
Escalator and Person move in Opposite Direction – [M – E]
Under “Escalators” the distance is constant. So this results in, $$D = Constant \longrightarrow S \propto \frac{1}{T}$$
Number of Steps in an Escalator (N) is always constant and can be calculated as – $$No. \ of \ Steps \ by \ person + No \ of \ Steps \ by escalator$$
Example – Person = 2 steps/sec , Escalator = 1 step/sec and Time taken to reach the 1st Floor = 30 seconds
N = [(2+1)*30] = (3*30) = 90 steps
Clocks
Hour hand makes an angle of 30 degrees in 1 hour.
Minute hand makes an angle of 30 degrees in 5 minutes.
Hour hand covers a distance of 5 minutes in 1 hour.
1 minute contains 6 degree angle..
In each hour, minute hand covers 55 minutes distance more than hour hand.
When both minute hand and hour hands makes an angle of 90 degrees then both lie 15 minutes apart from each other.
HOUR HAND
MINUTE HAND
SECOND HAND
$$12 \ hr \rightarrow 360^\circ $$
$$60 \ min \rightarrow 360^\circ $$
$$60 \ sec \rightarrow 360^\circ $$
$$1 \ hr \rightarrow 30^\circ $$
$$1 \ min \rightarrow 6^\circ $$
$$1 \ sec \rightarrow 6^\circ $$
Relative Speed of Hour and minute Hand – $$Relative \ Speed = {\LARGE [} 6 – \frac{1}{2} {\LARGE ]} = {\LARGE[}5 \frac{1}{2} {\LARGE]}^\circ {\LARGE/} minute $$
Similarly, Hour and Minute Hand meet at 11 Equidistant points within 12 hours and the ratio of their speeds are – $$Hour \ Hand : Minute \ Hand = 1 : 12 \ [Ratio \ of \ Speed]$$
Interval of Each Equidistant Gap – After every “66(5/11) minutes” in a clock, the hour hand and minute hand will meet or coincide with each other $$12 \ hours \rightarrow 11 \ Equidistant \ Gaps \\ 1 \ Equidistant \ Gap \rightarrow \frac{12}{11} * 60 = \frac{720}{11} mins = {\LARGE[} 65 \frac{5}{11} mins {\LARGE]}$$
Angle between Hour Hand and Minute Hand – $$\theta = Angle \ travelled \ by \ Hour \ Hand – Angle \ travelled \ by \ Minute \ Hand$$ $$\theta = {\LARGE[}30hr – \frac{11}{2}min {\LARGE]} \ {\Large or} \ {\LARGE[} \frac{11}{2}min – 30h {\LARGE]} \\ [When \ hour \ hand \ above \ minute \ hand] \ {\Large or} \ [When \ minute \ hand \ above \ hour \ hand]$$
A clock makes 2 distinct right angles between any 2 hours, but a clock makes 44 distinct right angles between 24 hours (not 48 because a few coincide)
Application of Relative Speed in Clocks –
In 1 minute, Minute hand covers “6 degrees” and Hour hand covers “1/2 degrees”, so $$Relative \ Speed = {\LARGE [} 6 – \frac{1}{2} {\LARGE ]} = {\LARGE[}5 \frac{1}{2} {\LARGE]}^\circ {\LARGE/} minute $$
In 1 hour, Minute hand covers “60 minutes” divisions, so $$Relative \ Speed = [ 60 – 5 ] = 55 \ minute \ division \ per \ hour$$
