Today we’ll cover Time, Speed and Distance basic concepts. The type of questions varies from simple to quite complex. But No need to worry if you know the basics.
There are many formulas deducible based solely on the basic concepts. Today, we’ll try to know what these formulas says in general so that we don’t burden ourselves remembering them.
Time never stops!
If I’m typing all this, and I stop in the middle, then only my average speed of typing goes down. Time never stops. With each second that pass, Present becomes past and future becomes present. This is the first thing you need to know.
The second thing is speed.
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:
Quiz Ques 1: Uday travels one third of its journey by train with speed of 60kmph and the rest of journey by car with speed of 80kmph. Find his average speed of his journey?
(Answer to this and all quiz questions later)
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)
Quiz Ques 2: Can you solve further and find the increased speed of Rajeev?
Let’s now solve a very good question which will clear many concepts in a single run!!
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)
Ta/Tb = square of [40/30] = square of [4/3] = 16/9èSb = 40 kmph, Sa = 30kmph
(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.
Understood? Now, try to solve this question!!
Quiz Ques 3: A lives at P and B lives at Q. A usually goes to meet B at Q. He covers the distance in 3 hour at 150kmph. On a particular day, B started moving away from A While A was moving towards Q, thus A took 5 hours to meet B. What is the speed of B?
Concept of Boats and Stream
The concepts of boats and streams is also based on this relative speed.
When boat goes downstream, the speed of flowing water helps the boat to move faster with more speed. When boat goes upstream, the speed of flowing water tries to cancel the speed of boat. The boat moves slower this time.
If, speed of stream = S and speed of boat is B, then
Downstream speed, D = B + S
Upstream speed, U = B – S
B generally means speed of boat in still water.
Hence, B = [D+U]/2 and S = D-B = [D-U]/2
Let’s apply this concept in this question:
Ques: A man can row 9 kmph in still water. It takes him twice as long as to row up as to row down, Find the rate of stream of water.
Let distance covered by boat be ‘D’
Speed of stream be ‘S’
Speed of boat is 9kmph
- Downstream time Td = distance/speed = D/[9+S]
- Upstream time Tu = distance/speed = D/[9-S]
Tu is twice than Td
- D/[9-S] = 2*D/[9+S]
On solving, we will get S = 3 kmph
Understood? Solve this question now:
Quiz Ques 4: A man can row at 10 kmph in still water. If the river flows at 3 kmph and, it takes 12 hours more in upstream than to go downstream for the same distance. How far is the place?
Concept of Races
In races, questions are asked of two or three player race. Questions are like,
Ques: In a race between Ram and Rahim, Ram has won the 1 km race by 100 meters. What is the ratio of their speeds?
The concepts are no different than those that we have already covered. Just some twists in the questions. The objective is to find what the question is saying. Answers will follow.
Now, first step and the most important step is to feel in mind’s eye what all is happening in the race. When Ram has just won the race of 1km, Rahim is how far behind him?
Ans is 100 meters behind him.
And Ram has covered 1000 meters, Rahim has covered how much distance?
Ans is 900 meters.
And Ram has covered 1000 meters in the same time it took Rahim to cover 900 meters.
Distance covered are in the ratio of 1000:900
Know the previous concept that when distance is different and time is constant, speed is directly proportional to the distance.
So, speed ratio of Ram : Rahim will also be 1000:900 = 10 : 9
This question will cover the remaining logic.
Ques: in a 1000m race, Ravi gives Vinod a start of 40m and beats him by 19 seconds. If Ravi gives a start of 30 sec to Vinod, then Vinod beats Ravi by 40m. What is the ratio of speed of Ravi to that of Vinod?
Case1: Now, visualize, in 1000m race, Ravi has given Vinod a start of 40m and beats him by 19 seconds. Means to say, Ravi runs 1000m while Vinod runs only 960m.
Second thing, when Ravi completed his 1000m, Vinod is still running and he runs for 19s more.
When putting it mathematically, if Ravi has completed the 1000m in T1 seconds, Vinod took T1+19 seconds to complete the 960m.
Case2: Ravi gives Vinod a start of 30s, then Vinod beat Ravi by 40m. Means to say, When the race is finished, Vinod has run 1000m while Ravi has run only 960m.
Second thing, Vinod has given the start of 30s. When Vinod has completed his 1000m, Ravi is still behind him 40m (i.e. Ravi has completed 960m)
When putting it mathematically, if Vinod has run 1000m in T2+30 seconds, Ravi has run 960m in T2 seconds.
Based on all the above facts, we’ll find their speeds.
Ravi’s speed = 1000/T1 = 960/T2
T1 = [25/24]*T2
Also, Vinod’s Speed = 960/[T1 + 19] == 1000/[T2 + 30]
Solving this by putting T1’s value, we gets, T2 = 120s
Required ratio = [960/T2] / [1000/(T2 + 30)] == [960/120] / [1000/150] = 6/5 = Answer.
If you understood all this, try solvingthesequestion using the same basic concepts.
Quiz Ques5: In a 1600m race, A beats B by 80m and C by 60m. If they run at the same time, then by what distance will C beat B in a 400 m race?
Quiz Ques 6: A beats B by 100m in a race of 1200m and B beats C by 200m in a race of 1600m. Approximately by how many meters can A beat C in a race of 9600m?
Circular Motion Concept
In circular tracks, the radius should be given or the length of the track is given. If length is given then its okay, if radius is given then put the formula L = 2*pi*radius to find the length of track.
Two or more runners will run this track with unequal speeds. They will run in the same direction or opposite direction.
We’ve covered that when two people run in same direction, their relative speed = speed of person with more speed – speed of person with less speed.
And when they run in opposite direction, their relative speed becomes = speed of 1st person + speed of 2nd person.
This same concept is to be used here.
Time taken by them to meet for first time will be = length of trace / relative speed.
Means to say, if A runs 1000m race with speed of 25mps and B runs race with speed of 15mps and they both are running in opposite direction, then
Time taken = length of track 1000m / relative speed 25+15mps = 1000/40 = 25seconds
If they run in the same direction, they will take = 1000/10 = 100 seconds.
Sometimes, question is asked of their meeting at the same starting point. For, this, LCM of their time is taken
That is to say, if A runs 1000m with 25mps speed, he will take = 40sec
B runs 1000m with 15mps speed = he will take = 200/3 sec
LCM of 40 and 200/3 = 200
So, they will take 200 sec to meet again at starting point.