Speed, Time and Distance : Race
📝Topic: Race
✍️Introduction: He zirlai hi Central service bei tur te tan a pawimawh em em a,he Chapter ah hian Question chhuak lar zual te mai baka he lam hawi problem awm thei te solve dan kan en ho dawn a ni...Math ah chuan engtin nge kan solve ang tih hi a pawimawh ber.Chumi ti thei tur chuan a awmzia leh nihna te kan hriatthiam hmasak hi a tul fo thin a ni.
Heng zirlai ah hian hriatthiam loh nei kan awm chuan Private lam ah pawh an zawh theih reng bawk e..
@Copyright reserved by the author ( Elisa Lalrinngheta)
✅Race Scenarios:
1. Single Race Scenario:
Pakhat chiah a tlan a nih chuan, an tlan chin leh an tlan chakna time hman chu an chhut thei.
2. Multiple Participants:
Mi pahnih emaw a aia tam an tlan a nih chuan, an tlan chak dan a zirin tunge hmasa zawk a chhut thei.
Mi pahnih an tlan chak dan a dang a nih chuan, an time hman an chhut thei a, tunge chak zawk a hriat thei.
3. Relative Speed:
An intlan pelh dawn emaw, an in hmachhawn dawn emaw chuan, relative speed an hmang thin.
🕵️An intlan pelh dawn chuan:
Relative Speed = Tlan chak zawk speed - Tlan muang zawk speed
An in hmachhawn dawn chuan:
Relative Speed = Participant A speed + Participant B speed
Example 1: Tlan mi pahnih, A leh B, an tlan tan na hmun a inang a, an finish line pawh a inang. Runner A chu 8 km/h in a tlan a, Runner B chu 10 km/h in a tlan. Runner A chu 30 minutes a hmasa in a tlan a nih chuan, Runner B in a catch up hun time chu engtia rei nge?
Solution:
Runner A in 30 minutes a tlan chin chu:
In 30 minutes, Runner A covers:
Distance= speed x time = 8 x 0.5 = 4 km
Relative speed hriat: B chakna A nen a an relative speed:
🕵️Relative speed= 10 - 8 = 2 km/h
man hun:
Hun = Distance / Relative speed
Hun = 4 / 2 = 2 hours
. : A answer chu 2 hours a ni.
Entirna 2: Tlan tute pahnih, Mawia leh Liana, chu 200-meter intlansiakah an tel. Mawia chu second khata meter 9 chakin a tlan a, Liana chu second khata meter 7 chakin a tlan thung.
1.Tlan tute pahnihin intlansiak an zawh nan engtia rei nge an mamawh?
2.Mawi-an liana a hneh theih na engtia thui nge a mamawh?
[Example 2: Two runners, Mawia and Liana, participate in a 200-meter race. Mawia runs at a speed of 9 meters per second, while liana runs at a speed of 7 meters per second.
1. How long does it take for each runner to complete the race?
2. By how much distance does mawia beat liana when he finishes the race?]
Solutions: Given: Speed_a= 9 m, Speed_b= 7 m
A hmasa ber ah chuan an pahnih in intlansiak an zawh nan a hun an mamawh zat en hmasat tur:
Alex-a hun hman chhutna:
Hun (A) angin sawi ang (Time_A)= Hlat zawng / Chak zawng = 200 / 9 = 22.22 seconds.
✅ Liana-a tan:
Hun(B) (Time_B)= 200/7=28.57
seconds.
Mawia-an tlan tawp a thlen huna Liana hlat zawng (distance) chhutna:
Mawia-an tlankawng a zawh zawh hun chhung (22.22 seconds vel) hian, liana hian heti zat hi a tlan hman ang:
Hlat zawng (Distance_Ben)= speed_b × Time_a= 7×22.22= 155.56 meters.
Mawi-an liana a lehpelh hlat zawng chhutna:
Lehpelh hlat zawng: Tlan zawng zawng - Lian-a tlan tawh zat.
D= 200-155.56 = 44.44 meters.
Chuvangin, a chhanna dik chu 44.44 m a ni.
✅ Circular Track (Bial kawng)
Running in the Same Direction (Kawng khat zawha tlan) chuan :
Tlantute chu kawng khat zawha an tlan laiin, kawng hmun hrang hranga an intawh hmasak ber hun chhung:
✅Problems:
🕵️.Running in the Opposite Direction:
🕵️✅ Example hetiang case bik ho hi en leh ang:
✅ Time after the racers meet each other exactly at the starting point:
The time when racers meet again at the starting point on a circular track is the Least Common Multiple (LCM) of the individual times each racer takes to complete one full lap. This formula applies regardless of whether they are running in the same or opposite directions, provided they start simultaneously.
Solution:
✅🕵️.Mi pathum A, B, leh C chu L meters a sei, circular track-ah P, Q, leh R m/s speed ve vein, direction inangah an tlan kual a, track khawi hmunah pawh an intawk thei.
An intawhkhawm hun chhung:
• P is the speed of person A.
• Q is the speed of person B.
• R is the speed of person C
📝Problems:( SSC, Railway)
Q1.Walking at the rate of 4 kmph a man covers certain distance in 2 hrs 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in how many minutes ?
a) 35 min.
b) 45 min.
c) 40 min.
d) 50 min.
Q2.In a race of 200 metres, B can give a start of 10 metres to A, and C can give a start of 20 metres to B. The start that C can give to A, in the same race, is
a) 25 metres
b) 27 metres
c) 29 metres
d) 30 metres
Q3.In a race of 800 metres, A can beat B by 40 metres. In a race of 500 metres, B can beat C by 5 metres. In a race of 200 metres, A will beat C by
a) 1.19 metre
b) 1.27 metre
c) 12.7 metre
d) 11.9 metre
Q4.Two persons A and B are running. around a circular track of length 1200 m with respective speeds of 27 kmph and 36 kmph. They started at the same time from the same point and are running in the same direction. When will they meet for the first time on the track (in seconds)? SSC CHSL 14/08/2023 (2nd Shift)
(a) 420 (b) 240 (c) 480 (d) 360
Q5.Two runners, Sony and Mony, start running on a circular track of length 200 mat speeds of 18 and 24 km/h, respectively, in the same direction. After how much time from the start will they meet again at the starting point? SSC CGL 14/07/2023 (1st shift)
(a) 120 sec✅
(c) 100 sec
(b) 110 sec
(d) 90 sec
Solutions: Left to the reader
Q6.A beats B by 20 metres in a 100 metre race. If B beats C by 10 metres in a 100 metre race, by how many metres does A beat C in a 100 metre race?
Solution:
Let the length of the race = 100 m.
When A finishes 100 m, B covers 80 m (since A beats B by 20 m).
When B finishes 100 m, C covers 90 m (since B beats C by 10 m).
Now, when A covers 100 m, B covers 80 m.
When B covers 80 m, C covers:
C=(90/100)×80=72 m
So, when A covers 100 m, C covers 72 m.
Therefore, A beats C by:
100−72=28 metres
✅ Final Answer: A beats C by 28 metres.
Q7.In a 200 m race, A can beat B by 20 m and B can beat C by 10 m. By how many metres can A beat C in the same race?
Solution:
When A covers 200 m, B covers 180 m (A beats B by 20 m).
When B covers 200 m, C covers 190 m (B beats C by 10 m).
When A covers 200 m, B covers 180 m.
When B covers 180 m, C covers:
C=(190/200)×180=171 m
So, when A covers 200 m, C covers 171 m.
Therefore, A beats C by:
200−171=29 metres
✅ Final Answer: A beats C by 29 metres
Q8.In a 150 m race, A runs at 5 m/s and B at 4 m/s. If A gives B a start of 10 m, who will win and by how much?
Solution:
Distance for A = 150 m
Distance for B = 150 - 10 = 140 m
Time taken by A:
Time A =150/5=30 s
Time taken by B:
Time B =140/4=35 s
A finishes in 30 s, B in 35 s. So, A wins.
Distance covered by B in 30 s: Distance=4×30=120 m
So, when A finishes, B is at 120 m (from starting line), i.e., A beats B by:150−120=30 m
✅Final Answer: A wins by 30 metres.
Q9.In a 100 m race, A gives B a start of 10 m and still beats him by 5 seconds. If A runs at 10 m/s, what is B's speed?
Solution:
Distance for A = 100 m
Distance for B = 90 m
Time taken by A = 100/10=10 s
Time taken by B = 10+5=15 s
So, B's speed:
Speed B =90/15=6 m/s
✅ Final Answer: B's speed is 6 m/s.
Q10.A and B run a race of 200 m. A runs at 8 m/s, B at 6 m/s. How much start should A give B so that the race ends in a dead heat?
