Problems on Relative Speed  Time, Speed & Distance

Authored by Nitin Gupta, Founder, Director at AlphaNumeric.
When two objects are moving simultaneously with speeds S1 and S2, the speed of any object when observed from the other object’s perspective is called the relative speed. And it is distinct from S1 or S2.
E.g.: Consider yourself to be sitting in a moving train and another train passes your train. If the other train is in opposite direction to your train, the speed of the other train appears to be far far more than it actually is. And if the other train is in the same direction as yours, it just appears to be inching ahead of your train very very slowly.This observed/perceived speed is called as relative speed and is calculated as:
If two objects are moving with speeds S1 and S2, their relative speed is
S1 + S2, if they are moving in opposite direction and
S1 – S2, if they are moving in same direction.Relative speed is usually considered when one has to find the time taken to meet or catch and it can be found as follows:
Time taken to meet/catch = Initial distance separating them/Relative Speed i.e. S1 ± S2A thief escapes from a prison at 2 pm and travels away at a speed of 30 kmph. The police realize the escape at 3:30 pm and start the chase then at a speed of 40 kmph. At what time will the police catch the thief? At what distance from the prison is the thief caught?
Approach 1 : Using Relative Speed
When the chase starts i.e. at 3: 30 pm, the thief has already run 30 × 1.5 = 45 kms.
Thus, this is the distance that separated the police and the thief and so the time taken to catch is 45/ (40  30) = 4.5 hours.
This 4.5 hours is measured from the time chase starts i.e. from 3:30 and thus the thief is caught at 8 pm.
The distance run by the police is 40 kmph × 4.5 hours = 180 kms
So the thief is caught at a distance of 180 kms from the prison.Approach 2: Using Proportionality
Method 1: Considering distance constant
Thief escapes from police station and say is caught at point X.
The police also run the same distance, from police station to X.
Thus, the distance run by police and thief is the same. Hence the time for which they are running will be inversely proportional to their speeds. Since ratio of speeds of police and thief is 4 : 3, time for which they are running will be 3 : 4. Also the difference in the time for which they are running is 1½ hour (thief starts running
1½ hours earlier). Thus, police and thief have been running for 3 × 1.5 = 4.5 hrs and 4 × 1.5 = 6 hrs respectively. And so the thief is caught at 3:30 pm + 4.5 hrs i.e. 8 pm (or 2 pm + 6 hrs i.e. 8 pm)
Method 2: Considering time as constant
If we consider the time interval for which the chase is on i.e. from the time police start chasing to the time thief is caught, then the time run will be the same. However in this case the distance run is different (because we are considering the event from 3:30 pm when thief has a headstart of 30 × 1.5 = 45 km).
With time being constant, the distances run will be proportional to speed. Thus, ratio of speed and distance will be 4 : 3. And we know that since 3:30, the police have to run a distance of 45 km more than the thief. Thus, 1 of ratio scale corresponds to 45 km and police will cover a total of 4 × 45 = 180 km in catching the thief. Time taken for this will be 180/40 = 4.5 hrs.Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30 am and travels at 50 km per hour towards Baroda situated 100 kms away. At 7:00 am Howrah  Ahmedabad Express leaves Baroda towards Ahmedabad and travels at 40 kms per hour. At 7:30 Mr. Shah, the traffic controller at Baroda realises that both the trains are running on the same tack. How much time does he have to avert a headon collision between the two trains? [cat 1999]
a) 15 minutes
b) 20 minutes
c) 25 minutes
d) 30 minutesConsidering the reference time to be 7:30 am, we would need to find the distance between the two trains at 7:30 am. From 6:30 to 7:30, the Navjivan Express would have travelled 50 kms and from 7 to 7:30, the HowrahAhmedabad would have travelled 20 kms.
Thus the distance separating the two trains at 7:30 would be 100 – 50 – 20 = 30 kms.Approach 1: Using Relative Speed
Time to meet (collide) from 7:30 am is 30 /(50 +40) = 1/3 hours = 20 minutes.
Approach 2: Using proportionality
Since 7:30 to the time they meet, the two trains are travelling for same time. Thus, the distances covered will be in the same ratio as their speeds i.e. 5 : 4. And we know that the total distance covered by the two trains in this time is 30 km.
Thus the trains individually cover, 5/9 * 30 = 50/3 km & 4/9 * 30 = 40/3km.
The time taken for this is (50/3)/50 = 1/ 3 hr = 20 minutesA train X departs from station A at 11:00 am for station B, which is 180 km away. Another train Y departs from station B at 11:00 am for station A. Train X travels of an average speed of 70 kmph and does not stop anywhere until it reaches station B. Train Y travels at an average speed of 50 kmph, but has to stop for 15 minutes at station C, which is 60 kms away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to the point where the trains cross each other? (CAT 2001)
a. 112
b. 118
c. 120
d. None of theseMethod1: train Y stops for 15 min : So in 15 min 'Y' can travel 50 * 15/60 = 12.5 km.
So now we can say that "y" start 12.5 km behind station B. so now the initial distance between them 192.5km. now both trains are starting at the same time ===> whenever they meet time is constant ===> so ratio of distance is 7:5.
distance travelled by "X" = 7/12 * 192.5 = 112 kmMethod2: 60/50 = 1.2 or 1 hr 12 minutes + 15 minutes = 1 hr 27 minutes
in this mean time train X will travel 70+18.9 = 98.9
now distance between them = 21.xx km
21/(70 + 50) = 0.175
0.175 * 70 = 12.25
total distance from A = 98.9 + 12.25 = 111.xx or 112 km approxA car is climbing up a hill at a speed of 90kmph. A rabbit, sitting at the top of the hill spots the car when it is 1170 km away, and start running down the hill towards the car at a speed of 140kmph. As soon as it meets the car, the rabbit turns back towards the top of the hill at a speed of 120kmph. The rabbit continues this to & from motion from the top of the hill towards the car and again back at the top of the hill till the car raches the top of the hill. Find the total distance covered by the rabbit.
The car will reach to the top of the hill after 1170/90 = 13 hrs.
Let total time during which the rabbit climbed downhill & uphill are d & u hrs respectively. Obviously , during every to & fro motion, the distance covered downhill is equal to distance covered uphill.
Therefore, 140d = 120u
==> d/u = 6/7 , also d + u = 13 ==> d= 6 & u= 7
Total distance covered by rabbit = 14d + 120u = 840 + 840 = 1680km.Two trains 150 miles apart are travelling toward each other along the same track. The first train goes 60 miles per hour; the second train rushes along at 90 miles per hour. A fly is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until they collide. If the fly's speed is 120 miles per hour, how far will it travel? find the distance travelled by fly in the direction of 1st train & in the direction of 2nd train.
First part : We want to know the total distance that the fly covers, so let's use Distance = Rate * Time to solve the problem. We already know the fly's rate of flight. If we can find the time that the fly spends in the air, we can figure out how far it travels.
Ignore the fly for a minute, and concentrate on the trains. The first train is traveling at 60 miles/ hour and the second train is going 90 miles/ hour, so they are approaching each other at 60 miles/ hour + 90 miles/ hour = 150 miles/ hour. Now we know the rate at which the trains are closing in on each other and their distance apart (150 miles), so we can find the time until they crash:
Distance = Rate * Time
Time = Distance / Rate
= (150 miles) / (150 miles/ hour)
= 1 hour.
The fly spends the same amount of time traveling as the trains. It goes 120 miles/ hour, so in the one hour the trains take to collide, the fly will go 120 miles.now try to find distance travelled in the direction of first train & distance travelled in the direction of 2nd train
Approach 1: they are meeting after 1hr ===> first train has travelled a distance of 60 miles. since fly has started from 1st train  it has also covered those 60 . but total distance travelled by fly is 120 ==> fly has travelled 60 extra which will be equally distributed in the direction of 1st & 2nd train (as to & fro) ===> in the direction of first train = 60 +30 = 90 & in the direction of 2nd train 30.
Approach 2: let forward direction is F & backward direction is B ==> always remember F:B = (f+t1)/(ft1) where f = speed of fly & t1 is the speed of train from where fly has started initially . here f:b = (120+60):(12060)= 3:1 . so divide 120 miles in the ratio 3:1==> 90 & 30.
Approach 3: 150/150 = 1 hr
in 1 hour fly will travel = 120 miles
forward distance = x
backward distance = y
x + y = 120
x  y = 60
2x = 180
x = 90 = forward distance
y = 30 = backward distanceSimilar question (Credits : Abhishek Jain)
There is a robbery in a shop and when the police reaches the destination, the thief is 5Km away. The speed of the thief is 10m/s and that of the police is 20m/s. Also, the police has a dog which continuously runs between the police and the thief to give the police information about the thief.
a) Find the total distance covered by the dog.
b) Find the distance covered by the dog in the forward direction.
(The speed of the dog is 30m/s)a) relative speed = 36
5/36 * 108 = 15b) forward distance = X
backward distance = Y
X + Y = 15
X  Y = 10
solving X = 12.5 km
proof of X  Y = 10
look try to visualize things
M(Q)(Police)(O)(Dog)+(thief)
let police and dog both are at M dog runs (Q+O) km and touch thief in this mean time police also ran Q km
now DOG will run backward
M(Q)1st point(S)(dog+police)(D)thief
say dog catch police in Z minutes in this mean time police covers S km
distance covered by dog in forward direction = Q+O
distance covered by dog in backward direction = D
now what i have done is
(Q+O)  D should be equal to (Q+S) where as S+D = O
Q+O  O + S = Q+S
RHS = LHSA military column of 1 km length marches with a constant speed of 6 kmph. A courier from the back of the column is sent to the head of the column by bike to deliver a message. Arrived at the front, he instantly returns. When he returns again at the back, the column has travelled a distance of 1 km since the start of his errand. The courier drove with constant speed. How fast did the courier bike?
The relative speed of the courier (compared to the column) is v6 on the journey to the front, and v+6 on the way back. The distance travelled (relatively to the column) by the courier is in both cases 1 km, therefore the total time he is biking is 1 /(v  6) + 1/(v + 6)
The column travels 1 km in 10 minutes (1/6 hour), therefore
1 /(v  6) + 1/(v + 6) = 1/6
This can be written as a quadratic equation (v+6) + (v6) = (v6) × (v+6) / 6 with as only allowed solution v = 6 + 6 × sqrt(2) km/h (approximately 14.49 km/h).(Warning  Tough question)
Moreshwar and Ganesh started travelling towards each other from their hometowns, Hyderabad and Bangalore respectively. They met at point P in between for the first time. As soon as they met, they exchanged their cars (which could travel with their predefined speeds only) and turned back to travel towards their respective hometown cities. As soon as they reached their hometowns, they again started travelling back towards the other city and met at point Q for the second time. Note that after meeting at point P they did not meet each other before they reached their respective hometown cities. What was the ratio of their speeds such that the distance PQ was the highest?
a) 2 : 5
b) 2 : 1
c) 2 : 3
d) 5 : 6
e) 1 : 3Let the speed of Moreshwar and Ganesh be u and v respectively. Their relative speed was u + v and the time they took for the first meeting was d/(u + v) where d is the total distance between the two cities. Therefore, the distance of the point P from Hyderabad (H), where they met for the first time is
L(HP) = u * d/(u+v)  eq 1
Similarly, after meeting they exchange their cars and start travelling back to
their own cities. Now the condition given is that they would not meet each
other before reaching their respective cities. Let us say that u > v. (We will
get similar results and the same ratio even if we consider v > u) This means
that Ganesh would not overtake Moreshwar before Moreshwar reaches
Hyderabad.
Now time taken by moreshwar to reach Hyderabad = (u/v) * {ud/u+v}v
In this time the distance travelled by ganesh = [{ud/u+v}/v] * u = (u/v) * {ud/u+v}This distance is not more than the distance between the point P to Bangalore
and back to Hyderabad.
== (u/v) * {ud /u+v} < {vd/u+v} + dSolving this we get,
(u + v)(u − 2v) < 0
Therefore u has to lie between −v and 2v. But since we have assumed u > v,
the allowed interval for u is v to 2v, both inclusive.
Now we need to find the distance of the point Q from Hyderabad so that we
can find the distance PQ.
After leaving point P and meeting again at point Q they would have travelled a
distance of 2d and their relative speed would be same as (u + v).
Therefore time taken by them to meet again = 2d /(u+v)
In this time distance travelled by moreshwar = 2vd/(u+v)
This is same as the distance between point P and Hyderabad and the
distance between Hyderabad and point Q.
L(HP) +L(HQ) = 2vd/(u+v)eq2Now, l(PQ) = l(HP) − l(HQ)
Solving for l(PQ) from equations (i) and (ii) we get,
L(PQ)= 2d(uv) /(u+v)Looking at the expression, we can see that l(PQ) is an increasing function
with the value of u. Therefore given the range of u, we would see that the
value of l(PQ) is maximum when u = 2v.
Hence, option 2.Yash Chopra’s office is 10 km from his home, which he covers in 2 hours,. One day, as he was on his way to office, he passed his friend’s office who immediately pointed out that he had forgotten to wear his shoes. Yash Chopra therefore turned back, put on his shoes and set off the office. As the passed his friend’s office, she again pointed out that now he had forgotten his glasses. Yash chopra again went back, collected his glasses and reached the office 3 hours late. What is the distance between his home and friend’s office?
a. 3.75
b. 7.35
c. 8.5
d. Insufficient dataLet the distance between home and friend’s office is x. His speed = 10/2 =5 kmph. Extra time taken is 3 hrs. So the total extra distance covered in this time duration is 15 km. This distance is traveled because he goes from home to friends office and back twice.
Thus 4x = 15 or x = 3.75.Last one for your practice
Tripti and Deepti walk back and forth between the town hall and the county station at respective speeds of 2 kmph and 3 kmph. They start simultaneously  Tripti from the town hall and Deepti from
the county station .If they croses each other for the first time 60 minutes from the start, at what distance from the county station will they cross each other fro the fifth time?
a. 3 km
b. 4 km
c. 2.5 km
d. None of these