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Speed and distance concept


In time-distance problems, if we take Distance as a constant thing, then speed and time becomes variables. We change speed, and thereby we changes the time taken.

Suppose, Delhi to Agra is 120 km. And my motorcycle covers 40 km in one hour. So, how much time I will take to reach Agra?

Simple! 3 hrs. time.

But my friend’s car covers 60 km in an hour. He will take how much time?

Simple! 2 hrs. time.

Means to say, my friend will reach Agra 1 hour before me.

So, keeping the distance constant, we got two times for two speeds. The time taken is inversely proportional to speed.

Basic formula we used here for calculation of time taken is:

Time taken = Distance/Speed

And using this formula, we can calculate speed, or, distance, if two other things are known

Speed = Distance/Time

Distance = Speed * Time


Feel this in mind before we go further…

Let’s now entertain the concept of average speed!

Question: I travel half of my journey by Bus with speed of 60kmph and the rest half on my friend’s motorcycle with speed of 80kmph. What is my average speed of total journey?

Average speed is that speed which covers the total distance in the total time (that is, the total time taken to cover the distance if I go by variable speeds)

Average speed = Total distance / Total time taken

Now, here in this question, ‘speed’ is variable (means changing). Distance is taken constant. So, Time taken will also be variable depending upon the speeds.

Time = Distance/speed, T= D/S

Let total distance be 2D, so that for 1st speed we have half distance ‘D’, and for second speed we have second half distance ‘D’

S1 = 60kmph
S2 = 80kmph

So, we have

T1 = D/60
T2 = D/80

Now, average speed = total distance / total time

Total distance = distance for 1st time + Distance for 2nd time = 2D
Total time = D/60 + D/80 = [7D/240]

Average speed will be = [2D] / ( [7D/240] ) = 480/7 kmph

Let’s derive this formula

Let 1st speed (60) = X
Let 2nd speed (80) = Y

T1 = D/X
T2 = D/Y

Total Distance = 2D
Total time = D/X + D/Y = [Y+X]*D/[XY]

Average Speed will be = [2D] / ( [Y+X]*D/[XY] ) = [2XY]/[X+Y]

Note: This formula we have derived taking the distances for both the speed as equal. So, if in questions, distances varies, this formula will fail to be applicable.

If you can remember the formula, then its fine, but if not, it’s still is fine. Problem is to just find the average speed. Our suggestion is to stick to basic concepts.

Now let’s test you:


The distance of the college and home of Rajeev is 80km. One day, he was late by 1 hour than normal time to leave for college, so he increased his speed by 4kmph and thus he reached to college at the normal time. What is the changed speed of Rajeev?

To solve this question, first feel what the question is saying.

Distance is 80km. It is constant. Only speed is changed.

Now, let’s say his normal speed is X kmph, Then he will reach the college in 80/X hour time. (Equation 1)

With [X+4] speed, he will reach the college in 80/[X+4] hour time. (Equation 2)

Now the question says, he is late 1 hour but with X+4 speed, he reaches the college on time.

That means time in (Equation 1) must be 1 hour more than the time in (Equation 2)

The distance between two places P and Q is 700km. Two persons A and B started towards Q and P from P and Q simultaneously. The speed of A is 30kmph and speed of B is 40kmph. They meet at a point M which lies on the way from P to Q.

(i) How long will they take to meet each other at M?

(ii) What is the ratio of PM : MQ?

(iii) What is the distance MQ?

(iv) What is the extra time needed by A to reach at Q than to reach at P by B?

(V) What is the ratio of time taken by A and B to reach their respective destinations after meeting at M?

(vi) In how many hours will they be separated by only 560 km before meeting each other?

(vii) How long will it take to separate then by 280 km from each other when they cross M (time to be considered after their meeting)?


First of all we’ll make a diagram. Diagram making is important for solving questions like these. Diagrams makes us feel the question a little more clearly.


The concept of relative motion is entertained here.

By relative motion, we means the motion of one thing with respect to another thing. Suppose you’re sitting on the pillion seat behind the motorcycle of your friend who is driving the motorcycle at 40kmph, then the relative speed of you with respect to motorcycle (or your friend) will be zero because for your friend, you are not moving an inch. But with respect to a person selling ice cream in the corner shop, your relative speed will be 40kmph, because for him, you’re moving with a speed of 40kmph.

Now, you steal his ice cream, and get ahead. He also had a bike and he’s now driving his bike behind you with a speed of 50kmph. Will he catch you?

Of course, he will. Coz now, the relative speed of him is 10kmph more with respect to you. He will catch you sooner.

The concepts when put mathematically is this:

If two bodies A and B are moving with speed Sa and Sb, then relative speed will be 

Sa – Sb, if they’re moving in the same direction, and

Sa + Sb, if they’re moving in the opposite direction.


(i) Now apply this concept.

A and B, both are moving in opposite direction with speeds of 30kmph and 40kmph. So, their relative speed will be?

Ans: 30+40 = 70kmph. 

They will take how much time to reach at point M?

They will cover total Distance = 700km / with speed of 70 kmph = will take Time = 10 hour to reach at point M

Understand this before you go further to solve the rest of questions!

(ii) It took 10 hour by both of them to reach at M.

With speed 30kmph, A has covered 30*10 = 300km = PM
With speed 40kmph, B has covered 40*10 = 400km = MQ

Ratio of distances PM : MQ = 300 : 400 = 3 : 4

Note: know this that if time is taken constant, the ratio of distances will be equal to the ratio of their speed. (Just because distance = speed * time)


D1 = S1*T1
D2 = S2*T2

T1 = T2

——> D1/D2 = S1/S2

(iii) Distance MQ = 400 km

(iv) Time taken by A to reach Q = distance/speed = 700km/30kmph = 70/3 hour
Time taken by B to reach P = distance/speed = 700km/40kmph = 70/4 hour

Extra time taken by A will be —- 70/3 – 70/4 = 70/12 hour

Understand this before you go further!!

(v) When A has reached at point M, A has covered 300km (because PM = 300km) and B has covered 400km (because MQ = 400km). Now, A has to cover 400km more and B has to cover 300km more. So,

Time Ta taken by A to cover MQ = distance MQ/speed = 400/30 hour
Time Tb taken by B to cover PM = distance PM/speed = 300/40 hour

Ratio of their time = Ta/Tb = [400/30]/[300/40] = 16/9

If derived (just like we’ve solved), we will get to know that this ratio [Ta/Tb] of their time is the ratio of the reciprocal of squares of their speed.

Why not we derive this?

Suppose A travels X km with speed Sa and B travels 700-X km with speed Sb and reaches point M in time T.

Time taken to reach point M will be equal.

i.e. [X]/Sa = [700-X]/Sb
i.e. [700-X]/[X] = [Sb/Sa]

Now, for A, rest distance to cover is 700-A with speed Sa, and for B, rest distance to cover is X with speed Sb, they will take time Ta and Tb to reach their destinations.

Ta = [700-X]/Sa
Tb = [X]/Sb

So, ratio of their times will be = Ta/Tb = ([700-X]/Sa) / ([X]/Sb) = ([700-X]/[X]) * ([Sb/Sa])

But we know that [700-X]/[X] = [Sb/Sa]

So, putting this in equation, we gets, Ta/Tb = ([Sb/Sa]) * ([Sb/Sa])

i.e. Ta/Tb = square of [Sb/Sa]  —— (note Sb/Sa and not Sa/Sb)

Sb = 40 kmph, Sa = 30kmph  Ta/Tb = square of [40/30] = square of [4/3] = 16/9

(vi) They will be separated by only 560 Km if they have covered 700-560 = 140 km.

With relative speed of 70kmph, they will cover 140km in 2 hour. So, that means, after 2 hour, they will be separated by 560km

(vii) Again, after crossing at the point M, their relative speed still will be the same. I.e. they will cover 280km in 280/7 = 4 hour time.

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