Distance \(D\)
Path length travelled. Use metres with seconds or kilometres with hours unless converted.
11 SSC CGL · Mathematics
Time, Speed and Distance
Handwritten-style notes for SSC CGL Tier I and Tier II.
01MAP IT!
The Big Picture — Graphic Mind Map
Track distance along a timeline; whenever two moving objects interact, replace ordinary speed with relative speed.
TIME
SPEED
DISTANCE
SPEED
DISTANCE
1. Core Triangledistance, speed, time and units
2. Average Speedtotal distance divided by total time
3. Relative Motionmeeting, chasing and overtaking
4. Trains & Boatseffective distance and current-adjusted speed
5. Races & Trackshead starts, beats, laps and meetings
6. Time Shiftdelay, early arrival, stoppage and speed change
One triangle, many journeys
Distance is the accumulated result of speed acting for time. Split a changing journey into constant-speed legs and add distances or times as required.
\[\boxed{D=ST}\qquad\boxed{T=\frac DS}\qquad\boxed{S=\frac DT}\]
\[\text{gap closed}=\text{relative speed}\times\text{interaction time}\]
02ZERO TO EXPERT
The Foundation (Basics) — Complete Concept Build
Master compatible units and motion ratios before using specialized train, boat or race formulas.
ACore quantities and unit conversion
Speed \(S\)
Distance travelled per unit time.
\[S=\frac DT\]
Time \(T\)
\[T=\frac DS\]
Time units must agree with the speed denominator.
km/h to m/s
\[1\text{ km/h}=\frac5{18}\text{ m/s}\]
m/s to km/h
\[1\text{ m/s}=\frac{18}{5}\text{ km/h}\]
Time units
\[1\text{ h}=60\text{ min}=3600\text{ s}\]
\[1\text{ km}=1000\text{ m}\]
The unit bridge has reciprocal arrows
km/h is the larger numerical unit, so converting to m/s usually makes the number smaller. Reverse the fraction to come back.
BDirect and inverse ratios
Fixed time
\[T\text{ fixed}\Rightarrow D_1:D_2=S_1:S_2\]
Fixed speed
\[S\text{ fixed}\Rightarrow D_1:D_2=T_1:T_2\]
Fixed distance
\[D\text{ fixed}\Rightarrow S_1:S_2=T_2:T_1\]
Speed and time are inverse for the same distance.
Speed rises by \(x\%\)
\[\frac{T_{\text{new}}}{T_{\text{old}}}=\frac{100}{100+x}\]
\[\text{time decrease}\%=\frac{x}{100+x}\times100\%\]
Speed falls by \(x\%\)
\[\frac{T_{\text{new}}}{T_{\text{old}}}=\frac{100}{100-x}\]
\[\text{time increase}\%=\frac{x}{100-x}\times100\%\]
Equal distance product
\[S_1T_1=S_2T_2\]
This is the fastest relation for changed-speed timing.
Same road: faster means less time
For fixed distance, speed and time sit on opposite sides of a seesaw. Percentage changes are not numerically equal because the bases differ.
\[S\uparrow\Rightarrow T\downarrow\quad\text{but not by the same percentage}\]
CAverage speed — total distance over total time
Universal formula
\[\boxed{S_{\text{avg}}=\frac{\sum D_i}{\sum T_i}=\frac{\sum D_i}{\sum D_i/S_i}}\]
Equal distances at \(x,y\)
\[S_{\text{avg}}=\frac{2xy}{x+y}\]
The harmonic mean, not the arithmetic mean.
Equal times at \(x,y\)
\[S_{\text{avg}}=\frac{x+y}{2}\]
Distance ratio \(a:b\)
\[S_{\text{avg}}=\frac{a+b}{a/x+b/y}\]
Time ratio \(a:b\)
\[S_{\text{avg}}=\frac{ax+by}{a+b}\]
Stoppage included
Average journey speed uses total elapsed time, including stops, unless “running speed” is specifically requested.
Average the journey, not the speed labels
Speeds receive weights through either the distance travelled or the time spent at them. Identify which quantity is equal before choosing a shortcut.
DRelative speed — closing or separating a gap
Opposite directions
\[S_{\text{rel}}=u+v\]
The gap changes by the sum each unit time.
Same direction
\[S_{\text{rel}}=|u-v|\]
The faster object gains only the speed difference.
Interaction time
\[T=\frac{\text{initial effective gap}}{S_{\text{rel}}}\]
Freeze one object and watch the other
Relative speed converts a two-object motion problem into a one-object gap-closing problem.
\[\text{gap at time }t=D_0-S_{\text{rel}}t\]
EMeeting, overtaking and position
Approaching from distance \(D\)
\[T=\frac{D}{u+v}\]
Distances travelled before meeting are in speed ratio \(u:v\).
Chasing with lead \(D\)
\[T=\frac{D}{u-v}\quad(u\gt v)\]
Meeting point
\[D_A:D_B=u:v\]
Use ratio parts to locate the meeting without first finding time.
FTrains — identify the effective distance
| Crossing event | Effective distance | Relative speed | Time |
|---|---|---|---|
| Pole/person/tree | Train length \(L\) | Train speed \(v\) | \(L/v\) |
| Platform/bridge/tunnel length \(P\) | \(L+P\) | \(v\) | \((L+P)/v\) |
| Two trains, opposite directions | \(L_1+L_2\) | \(v_1+v_2\) | \((L_1+L_2)/(v_1+v_2)\) |
| Two trains, same direction | \(L_1+L_2\) | \(|v_1-v_2|\) | \((L_1+L_2)/|v_1-v_2|\) |
Crossing finishes only when the rear clears
A point object requires one train length. An extended object adds its own length. Two trains add both lengths regardless of direction.
\[\text{time}=\frac{\text{sum of lengths to clear}}{\text{relative speed}}\]
Train length from pole time
\[L=vt\]
Convert \(v\) to m/s if \(t\) is in seconds.
Platform length
\[P=v(t_{\text{platform}}-t_{\text{pole}})\]
Man moving relative to train
Use \(v_{\text{train}}+v_{\text{man}}\) if opposite, and \(|v_{\text{train}}-v_{\text{man}}|\) if same direction.
GBoats and streams
Still-water and stream speeds
Let boat speed in still water be \(b\) and stream speed be \(s\), with \(b\gt s\).
Downstream
\[v_d=b+s\]
Upstream
\[v_u=b-s\]
Recover boat speed
\[b=\frac{v_d+v_u}{2}\]
Recover stream speed
\[s=\frac{v_d-v_u}{2}\]
Equal-distance round trip
\[S_{\text{avg}}=\frac{2v_dv_u}{v_d+v_u}=\frac{b^2-s^2}{b}\]
The current is a moving walkway
Downstream, the current helps; upstream, it opposes. Always require \(b\gt s\) for genuine upstream progress.
\[\text{ground speed}=\text{boat speed relative to water}\pm\text{stream speed}\]
HRaces, head starts and “beats by” language
Same finish time
In a race of length \(L\), if \(A\) beats \(B\) by \(d\) metres:
\[v_A:v_B=L:(L-d)\]
Beats by \(t\) seconds
If \(A\) finishes in time \(T_A\), then \(B\) takes \(T_A+t\):
\[v_A:v_B=(T_A+t):T_A\]
Head start \(h\)
If \(A\) runs \(L\) while \(B\) runs \(L-h\) in equal time:
\[v_A:v_B=L:(L-h)\]
At the winner’s finish instant, compare distances
Both runners have travelled for the same time. Their distances are therefore proportional to speeds.
ICircular tracks and repeated meetings
Opposite directions
\[T_{\text{meet}}=\frac{L}{u+v}\]
Same direction
\[T_{\text{overtake}}=\frac{L}{|u-v|}\]
Return together to start
If lap times are \(t_1,t_2,\ldots\):
\[T_{\text{return}}=\operatorname{LCM}(t_1,t_2,\ldots)\]
Use rational-time LCM after converting to common units.
One lap is the relative gap
Starting together on a circular track, the next same-point interaction occurs when relative motion covers one full circumference.
\[\text{relative distance}=L\]
JDelays, early arrival and changed speed
Late at \(u\), early at \(v\)
If late by \(t_1\) at speed \(u\), early by \(t_2\) at speed \(v\), and \(v\gt u\):
\[D=\frac{uv(t_1+t_2)}{v-u}\]
Scheduled travel time
\[T_{\text{scheduled}}=\frac Du-t_1=\frac Dv+t_2\]
Time saved by speed change
\[\Delta T=D\left(\frac1u-\frac1v\right)\]
Distance from time difference
\[D=\frac{uv\Delta T}{|v-u|}\]
Required speed after delay
If distance \(D\) must be covered in available time \(T_{\text{new}}\):
\[v_{\text{required}}=\frac{D}{T_{\text{new}}}\]
Catch-up after late start
If slower object has lead time \(t_0\), initial lead is \(vt_0\); divide by relative speed to get catch time.
The scheduled instant sits between early and late
The two travel times differ by \(t_1+t_2\). Equate the same distance at both speeds to recover the route length.
KStoppages and effective speed
General effective speed
\[S_{\text{eff}}=S_{\text{run}}\frac{T_{\text{run}}}{T_{\text{run}}+T_{\text{stop}}}\]
Stops \(m\) minutes in each clock hour
\[S_{\text{eff}}=S_{\text{run}}\frac{60-m}{60}\]
Stops \(m\) minutes after each hour of running
\[S_{\text{eff}}=S_{\text{run}}\frac{60}{60+m}\]
The wording creates a different cycle.
Running speed is not journey speed
Draw a cycle containing a moving block and a stopped block. Distance is earned only during the moving block, but average speed uses the whole clock.
\[\text{distance per cycle}=S_{\text{run}}T_{\text{run}}\]
LDecision table and boundaries
| Pattern | Use | Critical check |
|---|---|---|
| One journey | \(D=ST\) | Match distance and time units |
| Average speed | Total distance ÷ total elapsed time | Do not average speed labels blindly |
| Meeting/chasing | Effective gap ÷ relative speed | Add for opposite; subtract for same direction |
| Train crossing | Sum lengths to be cleared | Convert speed to m/s for seconds/metres |
| Boats | Still-water speed \(\pm\) stream speed | Need \(b\gt s\) for upstream travel |
| Race/head start | Compare distances in equal time | Winner’s finish instant is the reference |
| Circular track | One lap ÷ relative speed | Direction changes sum vs difference |
| Delay/stoppage | Build actual travel-time equation | Include stopped time in journey average |
03SAVE TIME
Short Tricks & Magic Formulas
Use ratios and relative speed before substituting large distances or converting every quantity.
1Conversion and inverse-change bank
Speed units
\[\text{km/h}\times\frac5{18}=\text{m/s}\]
Speed up by \(x\%\)
\[\text{time falls by }\frac{x}{100+x}\times100\%\]
Speed down by \(x\%\)
\[\text{time rises by }\frac{x}{100-x}\times100\%\]
2Average-speed bank
Equal distance
\[\frac{2xy}{x+y}\]
Equal time
\[\frac{x+y}{2}\]
Distance ratio \(a:b\)
\[\frac{a+b}{a/x+b/y}\]
3Relative motion and trains
Meet
\[T=\frac D{u+v}\]
Catch
\[T=\frac D{|u-v|}\]
Train clears object
\[T=\frac{\text{sum of effective lengths}}{\text{relative speed}}\]
4Boats, races and tracks
Boat/stream recovery
\[b=\frac{v_d+v_u}{2},\qquad s=\frac{v_d-v_u}{2}\]
Beat by \(d\) metres
\[v_A:v_B=L:(L-d)\]
Next circular meeting
\[T=\frac L{u+v}\text{ or }\frac L{|u-v|}\]
5Delay and stoppage shortcuts
Time difference at two speeds
\[D=\frac{uv\Delta T}{|v-u|}\]
Early plus late
\[D=\frac{uv(t_1+t_2)}{v-u}\]
Journey speed with stops
\[S_{\text{eff}}=S_{\text{run}}\frac{T_{\text{run}}}{T_{\text{total}}}\]
SSC motion speed dashboard
Normalize units, mark the effective distance, choose ordinary or relative speed, build the clock, and check direction.
\[\text{units}\to\text{distance}\to\text{relative speed}\to\text{time}\to\text{direction}\]
04DON'T RUSH
The SSC / TCS Traps — Red Flags 🚩
Wrong options average speeds directly, use the wrong relative direction, or omit an object’s length.
Trap 1: arithmetic mean used for equal distances
- Average speed is total distance divided by total time.
- Equal distances use \(2xy/(x+y)\).
- Arithmetic mean works for equal time, not equal distance.
Trap 2: relative speeds added in a chase
- Same direction uses \(|u-v|\).
- Opposite directions use \(u+v\).
- Draw arrows before choosing the operation.
Trap 3: train length forgotten
- Crossing a pole uses one train length.
- Crossing a platform uses train plus platform length.
- Two trains require the sum of both lengths.
km/h mixed with seconds
Convert speed to m/s before pairing with metres and seconds. The factor is \(5/18\).
Still-water and stream speeds swapped
Boat speed is the average of downstream and upstream speeds; stream speed is half their difference.
Running speed used as average speed
Stops add time but no distance. Include stoppage time in total journey time.
05LOCK IT IN
Memory Hooks & Mnemonics
Turn motion formulas into road, river, train and clock pictures.
“Distance sits above speed × time”
Cover the unknown in the triangle to select multiplication or division.
“Five-eighteenths goes to metres per second”
Reverse to \(18/5\) when returning to kilometres per hour.
“Face-to-face add; chase subtract”
Opposite arrows add speeds; same-direction arrows use the difference.
“Rear must clear”
Train-crossing distance includes every length that must pass before the rear clears the object.
“Current helps down, hurts up”
Downstream is \(b+s\); upstream is \(b-s\).
“Stops eat time, not distance”
Effective speed uses the whole elapsed clock, including stationary intervals.
Core\(D=ST\)
Equal-distance average\(2xy/(x+y)\)
Meet/chasegap ÷ relative speed
Train crossingsum lengths ÷ relative speed
Boat speeds\(b\pm s\)
Stoppagedistance ÷ total time