SSC GD Time, Speed & Distance

• Understanding Time, Speed and Distance SSC GD Basics
• The Main Formula: Speed
• Unit Conversion (The Important Step)
• Average Speed: Finding the Middle Ground
• Relative Speed: When Things Move Together
• Train Problems: Length Matters
• Boat and Stream (Basic): Moving in Water

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Understanding Time, Speed and Distance SSC GD Basics

Welcome to the notes for Time, Speed and Distance SSC GD! This topic is very important for the SSC GD Constable exam. We will learn how things move and how fast they go.

• Distance: How far you travel. (Example: Going from your home to the market, maybe 5 kilometers).
• Time: How long it takes you to travel that distance. (Example: It took 1 hour).
• Speed: How fast you are moving. It tells us the distance covered in a certain amount of time.

If you run fast, your speed is high. If you walk slowly, your speed is low. We need to solve many ssc gd time speed and distance questions to master this topic.

The Main Formula: Speed

Speed, Time, and Distance are best friends. If you know two of them, you can always find the third one using the main formula.

The Core Formulas

• Speed Formula: Speed is Distance divided by Time. $$Speed=\frac{Distance}{Time}$$
• Distance Formula: Distance is Speed multiplied by Time. $$Distance=Speed\times Time$$
• Time Formula: Time is Distance divided by Speed. $$Time=\frac{Distance}{Speed}$$

Solved Examples: Basic Calculations

Example 1: Finding Speed

A car travels 100 kilometers in 2 hours. What is the speed of the car?

Solution:

1. First, let’s understand what we need to find. We need the Speed.
2. Now, write down the formula: $$Speed=\frac{Distance}{Time}$$
3. Put the numbers in. Distance is 100 km. Time is 2 hours. $$Speed=\frac{100\text{ km}}{2\text{ hours}}$$
4. Calculate it. 100 divided by 2 is 50.
5. So, the answer is 50 km/hr.

Example 2: Finding Distance

A boy runs at a speed of 10 km/hr for 3 hours. How much distance did he cover?

Solution:

1. First, let’s understand what we need to find. We need the Distance.
2. Now, write down the formula: $$Distance=Speed\times Time$$
3. Put the numbers in. Speed is 10 km/hr. Time is 3 hours. $$Distance=10\text{ km/hr}\times 3\text{ hours}$$
4. Calculate it. 10 multiplied by 3 is 30.
5. So, the answer is 30 kilometers.

Example 3: Finding Time

A train needs to cover 450 km. It travels at a speed of 90 km/hr. How much time will it take?

Solution:

1. First, let’s understand what we need to find. We need the Time.
2. Now, write down the formula: $$Time=\frac{Distance}{Speed}$$
3. Put the numbers in. Distance is 450 km. Speed is 90 km/hr. $$Time=\frac{450\text{ km}}{90\text{ km/hr}}$$
4. Calculate it. 450 divided by 90 is 5.
5. So, the answer is 5 hours.

Example 4: Using the Time, Speed and Distance SSC GD Formula

A cyclist covers 200 meters in 20 seconds. What is his speed in meters per second (m/s)?

Solution:

1. We need to find the Speed in m/s.
2. Formula is: $$Speed=\frac{Distance}{Time}$$
3. Put the numbers in. Distance is 200 meters. Time is 20 seconds. $$Speed=\frac{200\text{ m}}{20\text{ s}}$$
4. Calculate it. 200 divided by 20 is 10.
5. So, the answer is 10 m/s.

Unit Conversion (The Important Step)

In Time, Speed and Distance SSC GD problems, the units must match! You cannot use kilometers (km) with seconds (s). You must change them first.

The two main units for speed are:

1. Kilometers per hour (km/hr)
2. Meters per second (m/s)

Conversion Rules

• To change km/hr to m/s: Multiply the speed by $\frac{5}{18}$.
  • Why 5/18? Because 1 km = 1000 meters, and 1 hour = 3600 seconds. $\frac{1000}{3600}$ simplifies to $\frac{5}{18}$.
• To change m/s to km/hr: Multiply the speed by $\frac{18}{5}$.

Solved Examples: Unit Conversion

Example 1: km/hr to m/s

Change a speed of 72 km/hr into m/s.

Solution:

1. We want to go from km/hr to m/s. We must multiply by $\frac{5}{18}$.
2. Write down the calculation: $$Speed\text{ in m/s}=72\times \frac{5}{18}$$
3. Simplify the numbers. 72 divided by 18 is 4. $$Speed=4\times 5$$
4. Calculate it. 4 times 5 is 20.
5. So, the answer is 20 m/s.

Example 2: m/s to km/hr

A runner is running at 15 m/s. What is his speed in km/hr?

Solution:

1. We want to go from m/s to km/hr. We must multiply by $\frac{18}{5}$.
2. Write down the calculation: $$Speed\text{ in km/hr}=15\times \frac{18}{5}$$
3. Simplify the numbers. 15 divided by 5 is 3. $$Speed=3\times 18$$
4. Calculate it. 3 times 18 is 54.
5. So, the answer is 54 km/hr.

Example 3: Finding Distance with Mixed Units

A bus travels at 36 km/hr for 10 seconds. How much distance (in meters) did it cover?

Solution:

1. The units are mixed (km/hr and seconds). We must change 36 km/hr to m/s first.
2. Change speed: $$36\times \frac{5}{18}=2\times 5=10\text{ m/s}$$
3. Now we find the Distance. Formula: $Distance=Speed\times Time$.
4. Put the numbers in. Speed is 10 m/s. Time is 10 seconds. $$Distance=10\times 10$$
5. So, the answer is 100 meters.

Example 4: Finding Time with Mixed Units

How much time (in seconds) will a person take to cover 500 meters if his speed is 45 km/hr? These are common ssc gd time speed and distance questions.

Solution:

1. The units are mixed (meters and km/hr). We must change 45 km/hr to m/s first.
2. Change speed: $$45\times \frac{5}{18}$$
3. Simplify the fraction (divide 45 and 18 by 9): $$\frac{5\times 5}{2}=\frac{25}{2}=12.5\text{ m/s}$$
4. Now we find the Time. Formula: $Time=\frac{Distance}{Speed}$.
5. Put the numbers in. Distance is 500 m. Speed is 12.5 m/s. $$Time=\frac{500}{12.5}$$
6. Calculate it. 500 divided by 12.5 is 40.
7. So, the answer is 40 seconds.

Average Speed: Finding the Middle Ground

Average speed is not always the simple average of the speeds. It is the total journey divided by the total time taken. This is a key concept in Time, Speed and Distance SSC GD.

Average Speed Formula

$$\text{Average Speed}=\frac{\text{Total Distance Covered}}{\text{Total Time Taken}}$$

Solved Examples: Average Speed

Example 1: Simple Average Speed

A man travels 100 km in 2 hours and then travels another 150 km in 3 hours. What is his average speed?

Solution:

1. First, find the Total Distance. $$\text{Total Distance}=100\text{ km}+150\text{ km}=250\text{ km}$$
2. Next, find the Total Time. $$\text{Total Time}=2\text{ hours}+3\text{ hours}=5\text{ hours}$$
3. Now, use the Average Speed formula: $$\text{Average Speed}=\frac{250\text{ km}}{5\text{ hours}}$$
4. Calculate it. 250 divided by 5 is 50.
5. So, the average speed is 50 km/hr.

Example 2: Equal Distance, Different Speeds

A person goes from City A to City B at 20 km/hr and immediately returns from City B to City A at 30 km/hr. What is his average speed for the whole trip?

Solution:

1. When the distance is the same both ways, we use a special shortcut formula (harmonic mean). Let $S_1$ and $S_2$ be the speeds. $$\text{Average Speed}=\frac{2\times S_1\times S_2}{S_1+S_2}$$
2. Put the numbers in. $S_1=20$, $S_2=30$. $$\text{Average Speed}=\frac{2\times 20\times 30}{20+30}$$
3. Calculate the top and bottom parts: $$\text{Average Speed}=\frac{1200}{50}$$
4. Simplify the fraction. 1200 divided by 50 is 24.
5. So, the average speed is 24 km/hr.

Example 3: Finding Time First

A car covers the first half of a 400 km journey at 80 km/hr and the second half at 100 km/hr. Find the average speed.

Solution:

1. The total distance is 400 km. Each half is 200 km.
2. Find Time 1 (for the first 200 km): $T_1=\frac{200}{80}=2.5\text{ hours}$.
3. Find Time 2 (for the second 200 km): $T_2=\frac{200}{100}=2\text{ hours}$.
4. Find Total Time: $2.5+2=4.5\text{ hours}$.
5. Calculate Average Speed: $\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}=\frac{400}{4.5}$
6. Calculate the final speed: $400/4.5\approx 88.89\text{ km/hr}$.

Example 4: Three Equal Distances

A man covers three equal distances at speeds of 10 km/hr, 15 km/hr, and 30 km/hr. Find the average speed. These are tough time speed and distance ssc gd questions.

Solution:

1. Since the distances are equal, we can use the formula for three speeds: $$\text{Average Speed}=\frac{3}{\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3}}$$
2. Find the sum of the reciprocals (use LCM of 10, 15, 30, which is 30): $$\frac{1}{10}+\frac{1}{15}+\frac{1}{30}=\frac{3}{30}+\frac{2}{30}+\frac{1}{30}=\frac{6}{30}=\frac{1}{5}$$
3. Now put this back into the formula: $$\text{Average Speed}=\frac{3}{\frac{1}{5}}$$
4. When you divide by a fraction, you flip it and multiply: $$\text{Average Speed}=3\times 5$$
5. So, the average speed is 15 km/hr.

Relative Speed: When Things Move Together

Relative speed is used when two objects are moving at the same time. We look at how fast they move in relation to each other. This is very common in ssc gd time speed and distance questions.

Rules for Relative Speed

1. Moving in the Same Direction: If two cars are going the same way, the relative speed is the difference between their speeds. They catch up slowly. $$\text{Relative Speed}=S_1-S_2$$ (Subtract the smaller speed from the bigger speed)

2. Moving in Opposite Directions: If two cars are coming towards each other, the relative speed is the sum of their speeds. They meet very quickly. $$\text{Relative Speed}=S_1+S_2$$

Solved Examples: Relative Speed

Example 1: Opposite Direction

Car A is traveling at 60 km/hr and Car B is traveling towards it at 40 km/hr. What is their relative speed?

Solution:

1. They are moving in Opposite Directions, so we add the speeds.
2. Write down the calculation: $$\text{Relative Speed}=60\text{ km/hr}+40\text{ km/hr}$$
3. Calculate it.
4. So, the relative speed is 100 km/hr.

Example 2: Same Direction (Chasing)

A police officer is chasing a thief. The officer runs at 12 km/hr and the thief runs at 10 km/hr. What is the relative speed?

Solution:

1. They are moving in the Same Direction, so we subtract the speeds.
2. Write down the calculation: $$\text{Relative Speed}=12\text{ km/hr}-10\text{ km/hr}$$
3. Calculate it.
4. So, the relative speed is 2 km/hr. This is how fast the officer is actually closing the gap.

Example 3: Finding Time to Meet

Two cities are 200 km apart. A train leaves City X at 50 km/hr, and at the same time, a train leaves City Y at 30 km/hr. How long will it take for them to meet?

Solution:

1. They are moving towards each other (Opposite Direction). Find the Relative Speed. $$\text{Relative Speed}=50+30=80\text{ km/hr}$$
2. The total distance they need to cover together is 200 km.
3. Use the Time formula: $Time=\frac{Distance}{\text{Relative Speed}}$. $$Time=\frac{200\text{ km}}{80\text{ km/hr}}$$
4. Calculate it. 200 divided by 80 is 2.5.
5. So, they will meet in 2.5 hours.

Example 4: Finding Time to Catch Up

A thief starts running at 40 km/hr. A police car starts chasing him 30 minutes later at 50 km/hr. How long (in hours) will it take the police to catch the thief?

Solution:

1. First, find the head start distance the thief covered in 30 minutes (0.5 hours). $$\text{Head Start Distance}=40\text{ km/hr}\times 0.5\text{ hr}=20\text{ km}$$
2. The police need to cover this 20 km gap. They are moving in the Same Direction. Find the Relative Speed. $$\text{Relative Speed}=50-40=10\text{ km/hr}$$
3. Use the Time formula: $Time=\frac{\text{Distance to Cover}}{\text{Relative Speed}}$. $$Time=\frac{20\text{ km}}{10\text{ km/hr}}$$
4. Calculate it. 20 divided by 10 is 2.
5. So, it will take the police 2 hours to catch the thief after the police started driving.

Train Problems: Length Matters

Train problems are special Time, Speed and Distance SSC GD problems because trains have length. When a train crosses something, the distance it covers is not just the distance to the object, but also its own length.

Distance Rules for Trains

1. Crossing a Point Object (Pole, Man, Signal): The object has no length. $$\text{Distance Covered}=\text{Length of the Train}(L_T)$$
2. Crossing a Long Object (Bridge, Platform, Tunnel, Another Train): $$\text{Distance Covered}=\text{Length of the Train}(L_T)+\text{Length of the Object}(L_O)$$

Remember to convert units to m/s if the answer needs to be in seconds!

Solved Examples: Train Problems

Example 1: Crossing a Pole

A 100-meter long train is traveling at 36 km/hr. How long will it take to cross a pole?

Solution:

1. First, convert the speed to m/s because the distance is in meters. $$\text{Speed}=36\times \frac{5}{18}=10\text{ m/s}$$
2. When crossing a pole, the Distance is the length of the train (100 m).
3. Use the Time formula: $Time=\frac{Distance}{Speed}$. $$Time=\frac{100\text{ m}}{10\text{ m/s}}$$
4. Calculate it. 100 divided by 10 is 10.
5. So, the time taken is 10 seconds.

Example 2: Crossing a Platform

A train 150 meters long is running at 54 km/hr. How long will it take to cross a 250-meter long platform?

Solution:

1. First, convert the speed to m/s. $$\text{Speed}=54\times \frac{5}{18}=3\times 5=15\text{ m/s}$$
2. Find the Total Distance (Train length + Platform length). $$\text{Distance}=150\text{ m}+250\text{ m}=400\text{ m}$$
3. Use the Time formula: $Time=\frac{Distance}{Speed}$. $$Time=\frac{400\text{ m}}{15\text{ m/s}}$$
4. Calculate it. 400 divided by 15 is approximately 26.67.
5. So, the time taken is 26.67 seconds.

Example 3: Crossing Another Train (Opposite Direction)

Train A (120 m long) runs at 40 km/hr. Train B (80 m long) runs towards it at 60 km/hr. How long will they take to cross each other? These are important time speed and distance ssc gd questions.

Solution:

1. Find the Total Distance (sum of both lengths). $$\text{Distance}=120\text{ m}+80\text{ m}=200\text{ m}$$
2. Find the Relative Speed (Opposite Direction, so add speeds). $$\text{Relative Speed}=40+60=100\text{ km/hr}$$
3. Convert Relative Speed to m/s. $$\text{Speed}=100\times \frac{5}{18}=\frac{500}{18}\approx 27.78\text{ m/s}$$
4. Use the Time formula: $Time=\frac{200}{\frac{500}{18}}$. $$Time=200\times \frac{18}{500}=\frac{3600}{500}=7.2$$
5. So, the time taken is 7.2 seconds.

Example 4: Crossing Another Train (Same Direction)

A faster train (150 m long) runs at 80 km/hr. A slower train (100 m long) runs in the same direction at 60 km/hr. How long does the faster train take to pass the slower train?

Solution:

1. Find the Total Distance (sum of both lengths). $$\text{Distance}=150\text{ m}+100\text{ m}=250\text{ m}$$
2. Find the Relative Speed (Same Direction, so subtract speeds). $$\text{Relative Speed}=80-60=20\text{ km/hr}$$
3. Convert Relative Speed to m/s. $$\text{Speed}=20\times \frac{5}{18}=\frac{100}{18}\text{ m/s}$$
4. Use the Time formula: $Time=\frac{250}{\frac{100}{18}}$. $$Time=250\times \frac{18}{100}=\frac{4500}{100}=45$$
5. So, the time taken is 45 seconds.

Boat and Stream (Basic): Moving in Water

This is a special type of Time, Speed and Distance SSC GD problem where the water itself is moving.

• Boat Speed (or Man Speed): Speed of the boat in still (calm) water ($S_B$).
• Stream Speed (or Current Speed): Speed of the water ($S_S$).

Key Concepts

1. Downstream (D): Moving with the water flow. The water helps the boat go faster. $$\text{Downstream Speed}(S_D)=S_B+S_S$$
2. Upstream (U): Moving against the water flow. The water slows the boat down. $$\text{Upstream Speed}(S_U)=S_B-S_S$$

Solved Examples: Boat and Stream

Example 1: Finding Downstream Speed

A boat travels at 10 km/hr in still water. The speed of the stream is 2 km/hr. What is the speed downstream?

Solution:

1. We are going Downstream, so the speeds add up.
2. Formula: $S_D=S_B+S_S$.
3. Put the numbers in. Boat speed is 10. Stream speed is 2. $$S_D=10+2$$
4. Calculate it.
5. So, the downstream speed is 12 km/hr.

Example 2: Finding Upstream Speed

A man can row at 8 km/hr in still water. If the river flows at 3 km/hr, what is his speed upstream?

Solution:

1. We are going Upstream, so the stream slows the man down (speeds subtract).
2. Formula: $S_U=S_B-S_S$.
3. Put the numbers in. Boat speed is 8. Stream speed is 3. $$S_U=8-3$$
4. Calculate it.
5. So, the upstream speed is 5 km/hr.

Example 3: Finding Speed of Boat in Still Water

A boat travels downstream at 15 km/hr and upstream at 9 km/hr. What is the speed of the boat in still water ($S_B$)?

Solution:

1. We know $S_D$ and $S_U$. We can use a shortcut formula to find the boat speed. $$\text{Boat Speed }(S_B)=\frac{S_D+S_U}{2}$$
2. Put the numbers in. $$S_B=\frac{15+9}{2}$$
3. Calculate the top part: $15+9=24$. $$S_B=\frac{24}{2}$$
4. Calculate it. 24 divided by 2 is 12.
5. So, the speed of the boat in still water is 12 km/hr.

Example 4: Finding Speed of the Stream

Using the data from Example 3 (Downstream 15 km/hr, Upstream 9 km/hr), what is the speed of the stream ($S_S$)? These are typical ssc gd time speed and distance questions.

Solution:

1. We know $S_D$ and $S_U$. We can use a shortcut formula to find the stream speed. $$\text{Stream Speed }(S_S)=\frac{S_D-S_U}{2}$$
2. Put the numbers in. $$S_S=\frac{15-9}{2}$$
3. Calculate the top part: $15-9=6$. $$S_S=\frac{6}{2}$$
4. Calculate it. 6 divided by 2 is 3.
5. So, the speed of the stream is 3 km/hr.