Time Speed & Distance
Concept:-
● Distance = Speed × Time
● Average Speed = \(\frac{{Total\;distance\;covered}}{{Total\;time\;taken}}\)
● A man covers a certain distance with S₁ km/h & come back with y km/h. He takes t hours to go and come back.
Now to find out distance:
Example: A boy goes to school at 3 km/h and return at a speed of 5 km/h. If he takes 9.6 hours in all. Find the distance of school from his home.
Sol:-
d = \(\frac{{3 \times 5}}{{(3 + 5)}} \times 9.6\)
d = 18 km Answer
● A man goes a certain distance with S₂ km/h & come back with S₁ km/h. If he takes t hours more to come back than go. Now to find the distance:
\(\frac{d}{{{S_2}}} – \frac{d}{{{S_1}}} = t\)
Example: A man goes to school at 3 km/h & comes back at a speed o 5 km/h. If he takes 4 hours more in Come back than go. Find the distance of school.
Sol:-
d = \(\frac{{3 \times 5}}{{(5 – 3)}} \times 4\) = 30 km Answer
● A man covers a certain distance at the speed of S₁ km/h, then he is late by ‘a’ hours but if he travels at the speed of S₂ km/h, then reaches ‘b’ hours earlier. Now to find the distance between two points:
Example: Starting from his house one day, a student walks @ 1.5 km/h and reaches his school 12 minute late. Next day he increases his speed @ 2 km/h and reaches the school 8 minutes earlier. How far is the school from his house ?
Sol:-
d = \(\frac{{1.5 \times 3.5}}{2} \times \frac{{\left( {20} \right)}}{{60}}\)
d = \(\frac{7}{8}\) Km Answer
Example: A man covers a certain distance by his car. Had he travelled 4 km/h faster then he takes 8 hour less time. But if he drive 3 km/h slower then he takes 18 hours more. Find the distance.
Sol:-
d = \(\frac{{S(S + 4)}}{4} \times 8 = \frac{{S \times (S – 3)}}{3} \times 18\)
⟹ S(S + 4)×2 = S(S – 3)×6
2S + 8 = 6S – 18
4S = 26
S = 6.5 km/h
∴ d = \(\frac{{6.5(6.5 + 4)}}{4} \times 8\)
d = 136.5 Km Answer
● Concept of relative speed:-
If speed has direction then it is called velocity i.e. velocity describes how fast and in what direction an object is moving
So if two objects have velocity V₁ & V₂ and distance between them be d then stop any one object & reverse direction of its velocity and give it to other object.
● Two trainds Tr₁ & Tr₂ are moving in the same direction with a speed x & y respectively (y > x). The driver of Train₁ observes that the Tr₂ coming from behind overtake and crossed his train completely in t₁ time whereas a driver on Tr₂ observer that he cross the Tr₁ in t₂ time. Then to find the ratio of length of both the train:
Tr₁ + Tr₂ = (y – x)t₁ …………… (i)
& Tr₁ = (y – x)t₂ ……………. (ii)
divide both equation
\(\frac{{T{r_1} + T{r_2}}}{{T{r_1}}} = \frac{{{t_1}}}{{{t_2}}}\)
\(1 + \frac{{T{r_2}}}{{T{r_1}}} = \frac{{{t_1}}}{{{t_2}}}\)
i.e. in such type of question there is not any need of speed of trains.
Example: The ratio of speeds of a goods and passenger train is 5 : 7 in the same direction. If passenger train crosses the goods train in 50 sec while a passenger in passenger train observes that he croses the goods train in 20 sec. Find the ratio of length of goods train to that of passenger train.
Sol: \(Lengt{h_{Goods\;:\;}}Lengt{h_{passenger}}\) = 20 : (50 – 20)
= 2 : 3 Answer
meeting time = ?
Let A to B distance = d
Speed of object₁ = S₁
Speed of object₂ = S₂
Let object meet after t time.
∴ distance travelled by obj₁ in t time + distance travelled by obj₂ in t time = total distance
S₁t + S₂t = d
t = \(\frac{d}{{{S_1} + {S_2}}}\) ………………. (i)
& d = S₁t₁ ⟹ S₁ = \(\frac{d}{{{t_1}}}\)
d = S₂t₂ ⟹ S₂ = \(\frac{d}{{{t_2}}}\)
put above value of S₁ & S₂ in equation (i)
t = \(\frac{d}{{\frac{d}{{{t_1}}} + \frac{d}{{{t_2}}}}}\)
Example: A train starts from delhi at 01:00 am and reached mumbai at 9:00 pm. Another train starts from mumbai at 1:00 am & reached delhi at 07:00 am next day. Find the meeting time ?
Sol: meeting time = \(\frac{{20 \times 30}}{{20 + 30}}\) = 12 h
= 01:00 am + 12 h
= 01:00 pm Answer
● 2 Guns were fired from the same place at an interval of ‘t’ hours but a man approaching the place hears the second firing after t₂ hours (t₂ < t₁) after the first. If sthen speed of sound is x km/h and that of man is y km/h, then ratio of speed of sound to that of man
Proof: the second gun was fired after t₁ hour. So if the man was not moving then he had heard its sound after t₂ hour
So, distance between place and man = xt₁ …………. (i)
Now, Since the man is moving towards Gun. So by using concept of relative speed ⟹ stop the man ⟹ reverse direction of its speed ⟹ give it to gun
then distance between place and man = (x + y)t₂ …………. (ii)
equate equation (i) and (ii)
xt₁ = (x + y)t₂
but if the man is going away from the gun
then equation (ii) become (x – y)t₂
Example: Two guns were fired from the same place at an interval of 28 minutes but a man sitting in the train approaching the gun hears the second firing after 26 minutes after th Iˢᵗ. If the speed of the sound is 325 m/sec, find the speed of train.
Sol:
\(\frac{{Speed\;of\;sound}}{{Speed\;of\;train}} = \frac{{26}}{{(28 – 26)}}\)
Example: The buses are departed after every 30 minutes from a bus stop but the man going away from the bus stop get the bus after every 35 minutes. Find the speed of Bus if the speed of man is 20 km/h ?
Sol: \(\frac{{Speed\;of\;bus}}{{Speed\;of\;man}} = \frac{{35}}{{(35 – 30)}}\)
● A and B starts walking towards each other. After meeting A covers his distance in t₁ time & B cover s his distance in t₂ time then ratio of their speeds
Let they meet after ‘t’ time
∴ At = Bt₂ ⟹ t = \(\frac{B}{A}{t_2}\) ……………….. (i)
& Bt = At₁ ⟹ t = \(\frac{A}{B}{t_2}\) ……………….. (ii)
equate equation (i) & (ii)
\(\frac{B}{A}{t_2} = \frac{A}{B}{t_1}\)
⟹ \(\frac{A}{B} = \sqrt {\frac{{{t_2}}}{{{t_1}}}} \)
(Q). 2 men A & B start moving from Delhi to jaipur at the same time toward each other, after meeting on the way they cover their remaining journey in 49 minute & 36 minute respectively. Find the faster speed if faster speed is 30 km/h more than slower.
Sol:-
● Without any stoppage a person travels a certain distance at avg. speed of F km/h and with stoppage it travels the same distance with 5 km/h avg. speed. To find how many min per hour does he stop
\(Minute/hour\;Stoppage = \left( {\frac{{Faster\;Speed\left( F \right) – Slower\;Speed\left( S \right)}}{{Faster\;Speed\left( F \right)}} \times 60} \right)\ \;min /h\)
Proof:- Let at faster speed F distance is covered in t₁ time & at slower speed S distance is covered in t₂ time.
∴ Ft₁ = St₂
\(\frac{{{t_1}}}{{{t_2}}} = \frac{S}{F}\) ⟹ \(\frac{{{t_1} – {t_2}}}{{{t_2}}} = \frac{{S – F}}{F}\)
Example: Excluding stoppage the speed of bus is 80 km/h & including stoppage it is 60 km/h. How many minute does the bus stops per hour.
Sol:- \(\left( {\frac{{80 – 60}}{{80}}} \right) \times 60\min /h\)
= 15 min/h Answer
t₂ – t = \(\left( {\frac{{d – {d_1}}}{v}} \right)\left( {\frac{{1 – a}}{a}} \right)\)
\(\left( {\frac{{d – {d_1}}}{v}} \right) = \left( {{t_2} – t} \right)\left( {\frac{a}{{1 – a}}} \right)\)
\(\frac{d}{v} – \frac{{{d_1}}}{v} = \left( {{t_2} – t} \right)\left( {\frac{a}{{1 – a}}} \right)\)
put value of \(\frac{d}{v}\) in above equation from equation (i).
\(\left( {\frac{{{t_1} – t}}{{1 – a}}} \right).a – \frac{{{d_1}}}{v} = \left( {\frac{{{t_2} – t}}{{1 – a}}} \right).a\)
\(\frac{{{d_1}}}{v} = \left( {\frac{a}{{1 – a}}} \right)\left( {{t_1} – {t_2}} \right)\)
Language of question: After travelling ‘x’ hours, a train meets with an accident. Due to this it has to stop for ‘t’ hours. After this the train starts moving at ‘a’ times of its original speed and reaches to its destination ‘t₁’ hour late. If the accident occured ‘d₁’ km ahead on the same line then the train reaches destination ‘t₂’ hour late. Find the original speed of the train.
Example: After travelling 5 hours a train meets with an accident. Due to this it has to stop for 2 hours. After this the train starts moving at \(55\frac{5}{9}\% \) of its speed, and reaches to its destination \(12\frac{2}{9}\) hours late. If the accident occured 150 km ahead on the same line then the train reaches destination \(10\frac{8}{9}\) hours late. Find the original speed of the train and distance travelled by the train.
Sol:-
Let speed of train = v
Normal time to reach form A to D = \(\frac{d}{v}\) ………………….. (i)
when accident happened at A & train travels at \(55\frac{5}{9}\% \) = \(\frac{5}{9}\) of its speed then
∴ subtract (i) from (ii)
Extra time taken to reach from A to D = \*2 + \frac{9}{{5v}}.d – \frac{d}{v}\)
⟹ \(12\frac{2}{9} = 2 + \frac{4}{5}.\frac{d}{v}\)
\(\frac{d}{v} = \frac{{115}}{9}\) ……………………… (iii)
when accident happened at B
Normal time to reach from B to D = \(\frac{{d – 150}}{v}\) …………………..(v)
Extra time to reach from B to D = \(2 + \frac{{d – 150}}{{\frac{5}{9}v}} – \frac{{d – 150}}{v}\)
= \(2 + \left( {\frac{{d – 150}}{v}} \right).\frac{4}{5}\)
⟹ \(10\frac{8}{9} = 2 + \left( {\frac{{d – 150}}{v}} \right).\frac{4}{5}\)
\(\frac{{d – 150}}{v} = \frac{{100}}{9}\)
⟹ \(\frac{d}{v} – \frac{{150}}{v} = \frac{{100}}{9}\)
put value of \(\frac{d}{v}\) in above equation from equation (iii
\(\frac{{115}}{9} – \frac{{150}}{v} = \frac{{100}}{9}\))
\(\frac{{150}}{v} = \frac{{115}}{9}\)
V = 90 km/h Answer
Put this value of V in equation (iii)
\(d = \frac{{115}}{9} \times 90 = 1150\;km\)
∴ total distance travelled = 5×90 + 1150
= 1600 Km Answer
method(2):
\(v = \frac{d}{{\left( {{t_1} – {t_2}} \right)}}.\left( {\frac{{1 – a}}{a}} \right)\)
\(v = \frac{{150}}{{\left( {12\frac{2}{9} – 10\frac{8}{9}} \right)}}.\left( {\frac{{1 – \frac{5}{9}}}{{\frac{5}{9}}}} \right)\)
= \(\frac{{150}}{{\frac{4}{3}}} \times \frac{4}{5}\)
= 90 km Answer
& total distance
= \(v.x + \left( {\frac{{{t_1} – t}}{{{t_1} – {t_2}}}} \right).{d_1}\)
= \(90 \times 5 + \left( {\frac{{12\frac{2}{9} – 2}}{{12\frac{2}{9} – 10\frac{8}{9}}}} \right).150\)
= 1600 km Answer
method(3):
∴ Speed = \(\frac{{150}}{{\frac{{115}}{9} – \frac{{100}}{9}}}\) = 90 km/h Ansewr
Total distance = 90×5 + \(\frac{{115}}{9}\)×90 = 1600 km Answer
To and Fro motion
Concept:
(a) When two bodies start moving towards each other
Let initial distance between them = D
Speed of A = \({S_A}\)
Speed of B = \({S_B}\)
● To meet they travel distances proportional to their individual velocities.
● They will cover the distance in proportion to their respective velocities regardless of the number of meetings. Thus total distance travelled for the nᵗʰ meeting is (2n – 1)D
(b) When two bodies start moving towards the same direction
● At any instance, the distance travelled by the bodies are proportional to the ratio of their velocities.
● Total distance travelled for nᵗʰ meeting is (2nD)