Given two variables, students are frequently asked to calculate the distance, speed, or travel time of something. These kinds of challenges are fun to solve because they reflect real-life circumstances for many people. A query might say, for example: Calculate the distance travelled by a car in 20 minutes at a constant speed of 50 km/hr. In most cases, we compute the desired quantity using the distance speed time formula. Let us consider some real-life problems.

Example 1: A car travels \(160 \mathrm{~km}\) in 4 hours. What is its speed in \(\mathrm{km} / \mathrm{hr}\)?

Solution:

We know the formula for speed is given by, Speed \(=\frac{\text { distance }}{\text { time }}\)
\(\Rightarrow\) speed \(=\frac{160}{4}=40 \mathrm{~km} / \mathrm{hr}\).
Therefore, the speed at which a car travels is \(40 \mathrm{~km} / \mathrm{hr}\).

Example 2: A truck was running from a city at an initial speed of \(40 \mathrm{kmph}\). The truck’s speed was increased by \(3 \mathrm{kmph}\) at the end of every hour. Find the total distance covered by the truck in the first 5 hours of the journey.

Solution: The total distance covered by the truck in the first 5 hours
\(
\begin{aligned}
& =40+43+46+49+52 \\
& =230 \mathrm{kms}
\end{aligned}
\)
Therefore, \(230 \mathrm{~km}\) is the total distance covered by the truck in the first 5 hours of the journey.

Average Speed

Average Speed \(=(\) Total distance traveled \() /(\) Total time taken \()\)

• Case-1: When the distance is constant: Average speed \(=2 x y / x+y\); Where, \(x\) and \(y\) are the two speeds at which the same distance has been covered.
• Case-2: When the time taken is constant: Average speed \(=(x+y) / 2 ;\) Where, \(x\) and \(y\) are the two speeds at which we traveled for the same time.

Example 3: A person travels from one place to another at 30 km/hr and returns at 120 km/hr. If the total time taken is 5 hours, then find the Distance.

Solution:

Here the Distance is constant, so the Time taken will be inversely proportional to the Speed. The ratio of Speed is given as 30:120, i.e. 1:4
So the ratio of Time taken will be 4:1. 
Total time taken = 5 hours; Time taken while going is 4 hours and returning is 1 hour. 
Hence, Distance = 30x 4 = 120 km

Example 4: Traveling at 3/4th of the original Speed a train is 10 minutes late. Find the usual Time taken by the train to complete the journey.

Solution:

Let the usual Speed be \(\mathrm{S} 1\) and usual Time be T1. As the Distance to be covered in both the cases is same, the ratio of usual Time to the Time taken when he is late will be the inverse of the usual Speed and the Speed when he is late

If the Speed is \(\mathrm{S} 2=3 / 4 \mathrm{~S} 1\) then the Time taken \(\mathrm{T} 2=4 / 3 \mathrm{~T} 1\)
Given \(\mathrm{T} 2-\mathrm{T} 1=10=>4 / 3 \mathrm{~T} 1-\mathrm{T} 1=10\)
\(\Rightarrow \mathrm{T} 1=30\) minutes.

Example 5: After traveling 50km, a train meets with an accident and travels at (3/4)th of the usual Speed and reaches 45 min late. Had the accident happened 10km further on it would have reached 35 min late. Find the usual Speed?

Solution:

Here there are 2 cases
Case 1: accident happens at \(50 \mathrm{~km}\)
Case 2: accident happens at \(60 \mathrm{~km}\)
The difference between two cases is only for the \(10 \mathrm{kms}\) between 50 and 60 . The time difference of 10 minutes is only due to these \(10 \mathrm{kms}\).
In case \(1,10 \mathrm{kms}\) between 50 and 60 is covered at \((3 / 4)\)th Speed.
In case 2, \(10 \mathrm{kms}\) between 50 and 60 is covered at the usual Speed.
So the usual Time \(t\) taken to cover \(10 \mathrm{kms}\), can be found out as below.
\(4 / 3 \mathrm{t}-t=10\) mins
\(=>t=30\) mins, \(d=10 \mathrm{kms}\)
so usual Speed \(=10 / 30 \mathrm{~min}=10 / 0.5=20 \mathrm{~km} / \mathrm{hr}\)

Example 6: Amit and Aman have to travel from Delhi to Jaipur in their respective cars. Amit is driving at 60 kmph while Aman is driving at 90 kmph. Find the Time taken by Aman to reach Jaipur if Amit takes 9 hrs.

Solution:

As the Distance covered is constant in both cases, the Time taken will be inversely proportional to the Speed. In the problem, the Speed of Amit and Aman is in a ratio 60: 90 or 2:3. So the ratio of the Time taken by Amit to that taken by Aman will be in the ratio 3:2. So if Amit takes 9 hrs, Aman will take 6 hrs.

Example 7: A man travels from his home to the office at 4km/hr and reaches his office 20 min late. If the Speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to the office.

Solution:

Let the Distance between home and office \(=d\). Suppose he reaches the office on Time, the Time taken \(=x\) minutes
Case 1: When he reaches the office 20 minutes late,
Time taken \(=x+20\)
Case 2: when he reaches the office 10 minutes early,
Time taken \(=x-10\) As the Distance traveled is the same, the ratio of Speed in case 1 to the Speed in case 2 will be the inverse of the Time taken in both cases Ratio of Speed in both cases \(=4: 6=2: 3\)
The ratio of Time in both cases \(=3: 2\)
Therefore \((x+20) /(x-10)=3 / 22 x+40=3 x\)
\(-30 x=70\) minutes
Taking case \(1,4=d /(90 / 60) \Rightarrow d=360 / 60=6 \mathrm{~km}\)

Example 8: Ram can row a boat in still water at \(10 \mathrm{kmph}\). He decides to go boating in a river. To row upstream he takes 2 hours and to row downstream, he takes 1.5 hours. Find the Speed of the river.

Solution:

Suppose the Speed of the river is ‘ \(y\) ‘ kmph.
While rowing upstream he takes \(2 \mathrm{hrs}\) and while rowing downstream he takes 1.5 hours.
As the Distance covered is constant the ratio of the net Speeds of the boat while going upstream and downstream will be the inverse of the ratio of the Time taken.
The ratio of Time taken (downstream: upstream) \(=1.5 / 2=3 / 4\)
So the ratio of the Speed of the boat (downstream: upstream) \(=4 / 3\)
Speed downstream: \(10+y\) Speed upstream : \(10-y\)
\(
(10+y) /((10-y))=4 / 3
\)
\(30+3 y=40-4 y\). Thus, \(7 y=10 \& Y=10 / 7\)
Speed of river \(=10 / 7 \mathrm{kmph}\)

Example 9: While going to office, Ramesh travels at a speed of 30 kmph and on his way back, he travels at a speed of 45 kmph. What is his average speed of the whole journey?

Solution:

When the distance travelled is the same, then average speed \(=2 a b /(a+b)\); (where \(a\) and \(b\) are two different speeds) therefore,

The Average Speed \(=\) Therefore, Average Speed \(=2 \times (45 \times 30) / (45+30)\), solving this we get \(36 \mathrm{kmph}\).

Example 10: A man walked at a speed of 4 km/hr from point A to B and came back from point B to A at a speed of 6 km/hr. What would be the ratio of the time taken by the man in walking from point A to B to that from point B to A?

Solution:

Ratio of speeds \(=4: 6=2: 3\).
\(\therefore \quad \text { Ratio of times taken }=\frac{1}{2}: \frac{1}{3}=3: 2 \text {. }\)