Distance = Speed * Time
Which implies →
Speed = Distance / Time and
Time = Distance / Speed
Let us take a look at some simple examples of distance, time and speed problems.
Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?
Time = Distance / speed = 20/4 = 5 hours.
Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.
Speed = Distance/time = 15/2 = 7.5 miles per hour.
Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?
Distance covered = 4*40 = 160 miles
Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph
Now, take a look at the following example:
Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?
Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.
Let us see how this question can be solved.
For these kinds of questions, a table like this might make it easier to solve.
Let the distance covered by that person be ‘d’.
Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’
IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.
He does this in a time of (d+7.5)/9.
Since the time is same in both the cases →
d/4 = (d+7.5)/9 → 9d = 4(d+7.5) → 9d=4d+30 → d = 6.
So, he covered a distance of 6 miles in 1.5 hours.
Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.
Here, we see that the distance is same.
Let us assume that its usual speed is ‘s’ and time is ‘t’, then
s*t = (1/3)s*(t+30) → t = t/3 + 10 → t = 15.
So the actual time taken to cover the distance is 15 minutes.
Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.
Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?
Let the time after which they meet be ‘t’ hours.
Then the time travelled by second train becomes ‘t-2’.
Distance covered by first train+Distance covered by second train = 320 miles
70t+20(t-2) = 320
Solving this gives t = 4.
So the two trains meet after 4 hours.
Example 2. A train leaves from a station and moves at a certain speed. After 2 hours, another train leaves from the same station and moves in the same direction at a speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed of the first train?
Let the speed of the first train be ‘s’.
Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours
Therefore, 6s = 60*4
Solving which gives s=40.
So the slower train is moving at the rate of 40 mph.
For problems with boats and streams,
Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream
[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]
Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream
[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]
Similarly, for airplanes travelling with/against the wind,
Speed of the plane with the wind = speed of the plane + speed of the wind
Speed of the plane against the wind = speed of the plane – speed of the wind
Let us look at some examples.
Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?
Let the distance be ‘d’ miles.
Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3
Speed upstream = 3-1 = 2 mph
Speed downstream = 3+1 = 4 mph
So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles.
Example 2. With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind?
Let the speed of the plane be ‘a’ and that of the wind be ‘w’.
Our table looks like this:
|With the wind||2400||a+w||4|
|Against the wind||2400||a-w||6|
4(a+w) = 2400 and 6(a-w) = 2400
Expressing one unknown variable in terms of the other makes it easier to solve, which means
a+w = 600 → w=600-a
Substituting the value of w in the second equation,
a-w = 400
a-(600-a) = 400 → a = 500
The speed of the plane is 500 kmph and that of the wind is 100 kmph.
Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.
Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.
The entire distance covered was 100 miles
So, 30t + 40(3-t) = 100
Solving which gives t=2.
Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.
Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.
d/30 + (100-d)/40 = 3
Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus.
Example 2. A plane covered a distance of 630 miles in 6 hours. For the first part of the trip, the average speed was 100 mph and for the second part of the trip, the average speed was 110 mph. what is the time it flew at each speed?
Our table looks like this.
|1st part of journey||d||100||t|
|2nd part of journey||630-d||110||6-t|
Assuming the distance covered in the 1st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.
Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.
From the first equation, d=100t
The second equation is 630-d = 110(6-t).
Substituting the value of d from the first equation, we get
630-100t = 110(6-t)
Solving this gives t=3.
So the plane flew the first part of the journey in 3 hours and the second part in 3 hours.
Example 2. Two persons are walking towards each other on a walking path that is 20 miles long. One is walking at the rate of 3 mph and the other at 4 mph. After how much time will they meet each other?
Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.
The time is ‘t’ for both of them because when they meet, they would have walked for the same time.
Since time is same, we can equate as
d/3 = (20-d)/4
Solving this gives d=60/7 miles (8.5 miles approximately)
Then t = 20/7 hours
So the two persons meet after 2 6/7 hours.
A boat covers a certain distance in 2 hours, while it comes back in 3 hours. If the speed of the stream is 4 kmph, what is the boat’s speed in still water?
A) 30 kmph
B) 20 kmph
C) 15 kmph
Let the speed of the boat be ‘s’ kmph.
Then, 2(s+4) = 3(s-4) → s=20
A cyclist travels for 3 hours, travelling for the first half of the journey at 12 mph and the second half at 15 mph. Find the total distance he covered.
A) 30 miles
B) 35 miles
C) 40 miles
D) 180 miles
Since it is mentioned, that the first ‘half’ of the journey is covered in 12 mph and the second in 15, the equation looks like
(d/2)/12 + (d/2)/15 = 3
Solving this gives d=40 miles