Speed, Distance and Time
Speed, Distance, and Time are important chapters for the purpose of the Maths section in aptitude exams. The basic concepts of Time, Speed, and Distance are used in solving questions based on straight-line motion, relative motion, circular motion, problems based on trains, problems based on boats, clocks, races, etc.
Let’s discuss them in detail about types of problems and solutions too!
Speed, Distance, and Time
Speed is the rate at which someone or something moves or operates or is able to move or operate.
Speed= Distance/ Time
It describes the distance traveled divided by the time taken to cover the distance.
Units: We can measure speed in different units like m/sec, km/hr, and in some cases, you will see km/min, m/min, feet/sec, and feet/hr.
Time is measured as the time taken by a process or activity that describes the distance traveled divided by the speed.
Time= Distance/ Speed
Units: We can measure time in different units like min, hour, and sec.
Distance can be defined as the length of space between two points. Distance is the product of speed and time.
Distance= Time * Speed
Units: We can measure distance in different units like km, meter, and miles.
Speed, Distance and Time conversions
- To convert from km / hour to m / sec, we multiply by 5 / 18. So, 1 km / hour = 5 / 18 m / sec
- To convert from m / sec to km / hour, we multiply by 18 / 5. So, 1 m / sec = 18 / 5 km / hour = 3.6 km / hour
- Similarly, 1 km/hr = 5/8 miles/hour
- 1 yard = 3 feet
- 1 kilometer= 1000 meters = 0.6214 mile
- 1 mile= 1.609 kilometer
- 1 hour= 60 minutes= 60*60 seconds= 3600 seconds
- 1 mile = 1760 yards
- 1 yard = 3 feet
- 1 mile = 5280 feet
- 1 mph = (1 x 1760) / (1 x 3600) = 22/45 yards/sec
- 1 mph = (1 x 5280) / (1 x 3600) = 22/15 ft/sec
- For a certain distance, if the ratio of speeds is a: b, then the ratio of times taken to cover the distance would be b: a and vice versa.
Now let’s explore some of the applications based on speed, distance, and time conversions.
Important Concepts and Definitions of Speed, Distance, and Time
Average speed is the mean value of the speed of a body over a period of time. The formula for average speed is needed since the speed of a moving body is not constant and varies across a period of time. Even with varying speed, the values of total time and the total distance covered can be used, and with the help of the formula for average speed, we can find a single value to represent the entire motion.
Average Speed = (Total distance traveled)/(Total time taken)
Case 1 – When the distance is constant: Average speed = 2xy/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.
Q1. A person travels from one place to another at 40 km/hr and returns at 160 km/hr. If the total time taken is 6 hours, then find the Distance.
Here the Distance is constant, so the Time taken will be inversely proportional to the Speed. The ratio of Speed is given as 40:160, i.e. 1:4
So the ratio of Time taken will be 4:1.
Total Time is taken = 6 hours; Time taken while going is 5 hours and returning is 1 hour.
Hence, Distance = 40x 5 = 200 km
Q.2. 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?
Let the usual Speed be S1 and the usual Time is T1. As the Distance to be covered in both the cases is the 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 S2 = ¾S1 then the Time taken T2 = 4/3 T1 Given T2 – T1 = 10 =>4/3 T1 – T1 = 10 => T1 = 30 minutes.
Inverse Proportionality of Speed & Time
Speed is inversely proportional to Time when the Distance is constant. S is inversely proportional to 1/T when D is constant. If the Speeds are in the ratio m:n then the Time taken will be in the ratio n:m.
There are two methods to solve questions:
- Using Inverse Proportionality
- Using Constant Product Rule
Q1. After traveling 60km, 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?
Using Inverse Proportionality Method
Here there are 2 cases
Case 1: accident happens at 60 km
Case 2: accident happens at 70 km
The difference between the two cases is only for the 10 km between 60 and 70. The time difference of 10 minutes is only due to these 10 km.
In case 1, 10 km between 60 and 70 is covered at (3/4)^th Speed.
In case 2, 10 km between 60 and 70 is covered at the usual Speed.
So the usual Time “t” taken to cover 10 km, can be found as below. 4/3 t – t = 10 mins = > t = 30 mins, d = 10 km
so usual Speed = 10/30min = 10/0.5 = 20 km/hr
Using Constant Product Rule Method
Let the actual Time taken be T
There is a (1/4) decrease in Speed, this will result in a (1/3)rd increase in Time taken as Speed and Time are inversely proportional
(A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other parameter if the parameters are inversely proportional) The delay due to this decrease is 10 minutes
Thus 1/3 T= 10 and T=30 minutes or ½ hour
Also, Distance = 10 km
Thus Speed = 20 kmph
NOTE: Let’s see the proportionality in the TSD(Time, Speed, Distance) equation:
- 1. s ∝ d if time is constant.
- 2. t ∝ d if speed is constant.
- 3. s ∝ 1/t if the distance is constant.
Let’s see an example of the above proportionality in the TSD equation.
A car moves for 4 hours at a speed of 25 kmph and another car moves for 5 hours at the same speed. Find the ratio of distances covered by the two cars.
Since the speed is constant, we can directly conclude that time ∝ distance. Hence = 𝑇𝑎 𝑇𝑏 𝐷𝑎 𝐷𝑏 Since the times of travel are 2 and 3 hours respectively, the ratio of distances covered is also 4/5.
Meeting Point Questions
If two people travel from two points A and B towards each other, and they meet at point P. The Total Distance covered by them at the meeting will be AB. The Time taken by both of them to meet will be the same. As the Time is constant, Distances AP and BP will be in the ratio of their Speed. Say that the Distance between A and B is d.
If two people are walking towards each other from A and B, When they meet for the First Time, they together cover a Distance “d” When they meet for the second time, they together cover a Distance “3d” When they meet for the third Time, they together cover a Distance of “5d”.
Q1. Harsh and Jai have to travel from Hyderabad to Bangalore in their respective cars. Harsh is driving at 60 kmph while Jai is driving at 90 kmph. Find the Time taken by Harsh to reach Bangalore if Jai takes 9 hrs.
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 Harsh and Jai is in ratio 60: 90 or 2:3.
So the ratio of the Time taken by Harsh to that taken by Jai will be in the ratio 3:2. So if Harsh takes 9 hrs, Jai will take 6 hrs.
In this concept, we will determine the movement and its relationships with respect to a moving point/body. In such situations, we have to take into account the movement of the body w.r.t. which we are trying to determine relative motion.
“Relation motion of a body is the motion of one body/point with respect to other body/point”
Case 1: Two cars C1 & C2 are moving in opposite directions. C1 moving at S1 kmph and C2 moving at S2 kmph. So, Relative speed S = S1+S2
Case 2: Two bodies are moving in the same direction. So, the Relative Speed S = S1 - S2
Q1 Two cars C1 & C2 are moving towards each other. C1 at 50 kmph and C2 at 30 kmph. The initial distance between them is 280 km. After how much time will they meet?
S1 = 50 kmph S2 = 30 kmph The speed with which they are approaching S = S1+S2 S = 50+30 = 80 kmph They have to approach each other and reach the meeting point. So, approaching distance/Relative distance= 280 km.
Hence, Relative Speed×Time = Relative Distance
80×t = 280 t = 3.5 hours. Therefore; they will meet after 3.5 hours.
Q2 Two cars C1 & C2 are moving in the same direction at a speed 50 kmph and 30 kmph respectively from the same point and they start moving at 2 pm. After how many hours will C1 be 140 km ahead of C2?
S1 = 50 kmph S2 = 30 kmph The Relative Speed S = S1-S2 S = 50 - 30 = 20 kmph Relative Distance = 140 km So, Relative Speed×Time = Relative Distance 20×t = 140 t = 7 hours. Therefore, after 7 hours C1 ahead 140 km from C2.
Frequently asked questions
What is the formula of time?
The formula for time is given as [Time = Distance ÷ Speed].
What is a speed-time graph?
In a speed-time graph, speed is always plotted on the vertical axis and time is always plotted on the horizontal. This represents the motion of a particle accelerating from a speed at time 0, u, to a speed v at time t.
What is the difference between speed and velocity?
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement. Put another way, speed is a scalar value, while velocity is a vector. In its simplest form, average velocity is calculated by dividing change in position (Δr) by change in time (Δt).
What is the speed-time curve?
A curve or line that best fits the points on a graph on which speed is plotted on the y-axis and time on the x-axis. A speed-time curve may be used to identify different levels of acceleration for example, during different phases of a 100-m race.
In this blog, we learned about Speed, Distance, and Time, with some examples. Then we learned about the most common concept around them, generally asked in aptitude exams viz. Average speed, Inverse proportionality, meeting points, relative speed. We also saw an example and sample problem for each concept to understand these concepts better.