11.1 Speed, Time & Distance Principles

11.1 Speed, Time & Distance Principles

Speed, Time, and Distance (STD) forms one of the most critical arithmetic modules in the SBI Clerk Preliminary and Main Examinations. In commercial banking operations, kinematic concepts directly model logistics: the scheduled transit of Currency Replenishment Vans (CRVs) transferring cash between Currency Chests and automated teller machines (ATMs), the routing of clearing instruments between regional collection centers, and the travel schedules of agricultural field officers inspecting Priority Sector Lending (PSL) crop assets across rural branches. In the Numerical Ability section, 2 to 3 questions test STD mechanics, either as stand-alone word problems or integrated into multi-variable Data Interpretation (DI) caselets.

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1. Fundamental Kinematic Relationships & Proportionality Axioms

The relationship connecting motion variables is defined by the core kinematic triad:

$$\text{Speed } (S) = \frac{\text{Distance } (D)}{\text{Time } (T)}, \qquad D = S \times T, \qquad T = \frac{D}{S}$$

In competitive banking examinations, solving problems within the 20-minute sectional window requires understanding how these variables interact under fixed physical constraints:

Axiom 1: Constant Distance (Inverse Proportionality of Speed and Time)

When the distance between origin and destination remains fixed, speed varies inversely with elapsed travel time:

$$D = \text{constant} \implies S \propto \frac{1}{T} \implies S_1 T_1 = S_2 T_2 \implies \frac{S_1}{S_2} = \frac{T_2}{T_1}$$

SBI Clerk Application: If an inspection officer increases his driving speed by $25\%$ (speed ratio changes from $4 : 5$), the time required to cover the same branch circuit decreases in the reciprocal ratio $5 : 4$ (a time reduction of $\frac{1}{5} = 20\%$).

Axiom 2: Constant Time (Direct Proportionality of Distance and Speed)

When two entities travel for the exact same duration, the distance traversed is directly proportional to their respective speeds:

$$T = \text{constant} \implies D \propto S \implies \frac{D_1}{D_2} = \frac{S_1}{S_2}$$

Axiom 3: Constant Speed (Direct Proportionality of Distance and Time)

When an entity moves at an unchanging velocity, distance traversed varies directly with elapsed time:

$$S = \text{constant} \implies D \propto T \implies \frac{D_1}{D_2} = \frac{T_1}{T_2}$$

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2. Metric Conversions & The Standard Speed Matrix

Banking exam problems frequently state velocities in kilometers per hour ($\text{km/h}$) while specifying obstacle lengths and transit intervals in meters ($\text{m}$) and seconds ($\text{s}$). Failure to align metric units immediately produces distractor answers.

Mathematical Derivation of Conversion Factors

1. Converting $\text{km/h}$ to $\text{m/s}$:

   $$1 \text{ km/h} = \frac{1 \text{ kilometer}}{1 \text{ hour}} = \frac{1,000 \text{ meters}}{3,600 \text{ seconds}} = \frac{10}{36} \text{ m/s} = \frac{5}{18} \text{ m/s}$$

   Rule: Multiply speed in $\text{km/h}$ by $\frac{5}{18}$ to obtain speed in $\text{m/s}$.

2. Converting $\text{m/s}$ to $\text{km/h}$:

   $$1 \text{ m/s} = \frac{1 / 1,000 \text{ km}}{1 / 3,600 \text{ h}} = \frac{3,600}{1,000} \text{ km/h} = \frac{18}{5} \text{ km/h} = 3.6 \text{ km/h}$$

   Rule: Multiply speed in $\text{m/s}$ by $\frac{18}{5}$ to obtain speed in $\text{km/h}$.

Standard Multiples Reference Table

Memorizing common multiples of $18 \text{ km/h}$ and $5 \text{ m/s}$ saves 20–30 seconds per calculation:

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3. Average Speed: Arithmetic Fallacy vs Harmonic Mean

A primary distractor in competitive exams is the simple arithmetic average $\frac{x + y}{2}$. In kinematic physics, average speed is defined strictly by the governing ratio:

$$\text{Average Speed } (S_{\text{avg}}) = \frac{\text{Total Distance Traversed}}{\text{Total Time Elapsed}} = \frac{D_{\text{total}}}{T_{\text{total}}}$$

Case A: Equal Distance Segments (Harmonic Mean)

When an entity travels a distance $D$ at speed $x$ and returns the same distance $D$ at speed $y$:

$$T_1 = \frac{D}{x}, \qquad T_2 = \frac{D}{y}, \qquad T_{\text{total}} = \frac{D}{x} + \frac{D}{y} = D \left(\frac{x + y}{xy}\right)$$ $$S_{\text{avg}} = \frac{2D}{D\left(\frac{x+y}{xy}\right)} = \frac{2xy}{x + y}$$

For three equal distance segments covered at speeds $x$, $y$, and $z$:

$$S_{\text{avg}} = \frac{3xyz}{xy + yz + zx}$$

Case B: Unequal Distance Segments

When a journey is partitioned into arbitrary distances $D_1, D_2, \dots, D_n$ covered at speeds $S_1, S_2, \dots, S_n$:

$$S_{\text{avg}} = \frac{D_1 + D_2 + \dots + D_n}{\frac{D_1}{S_1} + \frac{D_2}{S_2} + \dots + \frac{D_n}{S_n}}$$

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4. Relative Speed Principles

When two physical bodies are simultaneously in motion, their relative speed ($S_{\text{rel}}$) measures the rate at which the separation distance between them expands or contracts.

Opposite Direction (Approaching):  [Body A] -->  S1        S2  <-- [Body B]
                                   Relative Speed = S1 + S2

Same Direction (Overtaking):       [Body A] -->  S1    [Body B] -->  S2  (S1 > S2)
                                   Relative Speed = S1 - S2

1. Opposite Direction (Approaching or Diverging): Both speeds contribute to closing or widening the gap:

   $$S_{\text{rel}} = S_1 + S_2$$ $$\text{Time to Meet} = \frac{\text{Initial Separation Distance}}{S_1 + S_2}$$

2. Same Direction (Chasing or Overtaking, where $S_1 > S_2$): The faster body closes the gap at the net difference in velocities:

   $$S_{\text{rel}} = S_1 - S_2$$ $$\text{Time to Overtake} = \frac{\text{Initial Separation Distance}}{S_1 - S_2}$$

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5. Early and Late Arrival Dynamics (The Differential Shortcut)

A common question archetype presents an employee traveling to work at two different speeds, arriving late in one instance and early in another. Setting up dual linear equations consumes valuable exam time. Instead, use the Differential Distance Shortcut.

Derivation of the Shortcut

Let the scheduled travel time be $T_{\text{sched}}$ and the distance to the destination be $D$. Traveling at speed $S_1$ takes time $T_1 = \frac{D}{S_1}$, while traveling at speed $S_2$ takes time $T_2 = \frac{D}{S_2}$. The absolute difference between the two travel times is $\Delta t = |T_1 - T_2|$:

$$\left|\frac{D}{S_1} - \frac{D}{S_2}\right| = \Delta t \implies D \left(\frac{|S_2 - S_1|}{S_1 S_2}\right) = \Delta t$$ $$D = \frac{S_1 \times S_2}{|S_1 - S_2|} \times \Delta t$$

Determining the Net Time Difference ($\Delta t$)

To determine $\Delta t$, examine the arrival deviations relative to the scheduled time:

• Late by $t_1$ and Early by $t_2$: The deviations lie on opposite sides of the scheduled time $\implies \Delta t = t_1 + t_2$.
• Early by $t_1$ and Late by $t_2$: The deviations lie on opposite sides $\implies \Delta t = t_1 + t_2$.
• Late by $t_1$ and Late by $t_2$: Both lie on the same side $\implies \Delta t = |t_1 - t_2|$.
• Early by $t_1$ and Early by $t_2$: Both lie on the same side $\implies \Delta t = |t_1 - t_2|$.

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6. Symmetrical Post-Meeting Mechanics

Consider two travelers, A and B, who depart simultaneously from locations $X$ and $Y$ toward each other at uniform speeds $S_A$ and $S_B$. They pass each other at an intermediate meeting point $M$, after which traveler A takes $T_A$ hours to reach $Y$ and traveler B takes $T_B$ hours to reach $X$.

The Post-Meeting Theorem

$$\frac{S_A}{S_B} = \sqrt{\frac{T_B}{T_A}}$$

Mathematical Proof

Let $t$ be the elapsed time from departure until the instant of meeting at $M$. Because both travelers started simultaneously:

• Distance $X$ to $M = S_A \times t$
• Distance $M$ to $Y = S_B \times t$

After crossing at $M$, traveler A covers distance $M$ to $Y$ in $T_A$ hours, and traveler B covers distance $X$ to $M$ in $T_B$ hours:

$$M \text{ to } Y = S_A \times T_A, \qquad X \text{ to } M = S_B \times T_B$$

Equating expressions for the two segments:

$$S_B \times t = S_A \times T_A \implies t = \frac{S_A T_A}{S_B}$$ $$S_A \times t = S_B \times T_B \implies t = \frac{S_B T_B}{S_A}$$

Equating both expressions for $t$:

$$\frac{S_A T_A}{S_B} = \frac{S_B T_B}{S_A} \implies \frac{S_A^2}{S_B^2} = \frac{T_B}{T_A} \implies \frac{S_A}{S_B} = \sqrt{\frac{T_B}{T_A}}$$

Furthermore, the time elapsed until the meeting occurred is given by the geometric mean:

$$t = \sqrt{T_A \times T_B}$$

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7. Step-by-Step Worked Banking Logistics Example

Problem Statement:
An SBI Currency Replenishment Van (CRV) leaves the Currency Chest at 08:00 AM traveling at $40 \text{ km/h}$ to deliver fresh currency notes to a rural branch located $260 \text{ km}$ away. At 09:30 AM, a fast security escort jeep departs from the same Currency Chest along the identical route at $65 \text{ km/h}$ to intercept the van and provide high-value transit protection. Determine:

1. The exact time the escort jeep overtakes the cash van.
2. The distance from the Currency Chest at the point of interception.

Step 1: Calculate the Head Start Distance

The cash van travels alone from 08:00 AM to 09:30 AM (a duration of $1.5 \text{ hours}$):

$$\text{Head Start Distance} = 40 \text{ km/h} \times 1.5 \text{ h} = 60 \text{ km}$$

Step 2: Determine Relative Speed

Both vehicles travel in the same direction along the highway:

$$S_{\text{rel}} = S_{\text{jeep}} - S_{\text{van}} = 65 \text{ km/h} - 40 \text{ km/h} = 25 \text{ km/h}$$

Step 3: Compute Time Required to Overtake

$$\text{Time to Overtake} = \frac{\text{Separation Distance}}{S_{\text{rel}}} = \frac{60 \text{ km}}{25 \text{ km/h}} = 2.4 \text{ hours}$$

Converting 0.4 hours to minutes: $0.4 \times 60 = 24 \text{ minutes}$. Thus, transit time is 2 hours 24 minutes.

Step 4: Establish Clock Time and Interception Distance

• Clock Time: 09:30 AM $+ 2 \text{ hours } 24 \text{ minutes} = \mathbf{11:54 \text{ AM}}$.
• Interception Distance: $65 \text{ km/h} \times 2.4 \text{ h} = \mathbf{156 \text{ km}}$ from the Currency Chest.
  (Verification via van: $60 \text{ km} + (40 \text{ km/h} \times 2.4 \text{ h}) = 60 + 96 = 156 \text{ km}$. The van has traveled $156 \text{ km}$ of the $260 \text{ km}$ total journey, well before reaching the branch.)