Speed, Time, and Distance Concepts

1. Basic Speed-Distance-Time

• Formula:

$$\text{Distance}=\text{Speed}\times\text{Time}$$ $$\text{Speed}=\frac{\text{Distance}}{\text{Time}},\qquad \text{Time}=\frac{\text{Distance}}{\text{Speed}}$$

• Example: A car travels 180 km in 3 hours. Find its speed. Solution:

$$\text{Speed}=\frac{180}{3} =\boxed{60\text{ km/h}}$$

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2. Unit Conversion of Speed

• Formula:

$$1\text{ km/h}=\frac5{18}\text{ m/s}$$ $$1\text{ m/s}=\frac{18}{5}\text{ km/h}$$

• Example: Convert 72 km/h into m/s. Solution:

$$72\times\frac5{18} =\boxed{20\text{ m/s}}$$

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3. Relative Speed — Opposite Directions

• Formula:

$$\text{Relative speed}=v_1+v_2$$ $$\text{Time to meet}= \frac{\text{Initial distance}}{v_1+v_2}$$

• Example: Two cars are 300 km apart and travel towards each other at 60 km/h and 90 km/h. When will they meet? Solution:

$$v_{\text{relative}}=60+90=150\text{ km/h}$$ $$T=\frac{300}{150} =\boxed{2\text{ hours}}$$

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4. Relative Speed — Same Direction

• Formula:

$$\text{Relative speed}=|v_1-v_2|$$ $$\text{Time to overtake}= \frac{\text{Initial gap}}{|v_1-v_2|}$$

• Example: A car moving at 80 km/h is 20 km behind another car moving at 60 km/h. How long will it take to catch it? Solution:

$$v_{\text{relative}}=80-60=20\text{ km/h}$$ $$T=\frac{20}{20} =\boxed{1\text{ hour}}$$

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5. Trains Crossing a Pole

• Formula:

$$\text{Time}=\frac{\text{Length of train}}{\text{Speed}}$$

because the pole has negligible length.

• Example: A 180 m train moves at 54 km/h. How long does it take to cross a pole? Solution:

$$54\times\frac5{18}=15\text{ m/s}$$ $$T=\frac{180}{15} =\boxed{12\text{ seconds}}$$

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6. Train Crossing a Platform

• Formula:

$$\text{Time}= \frac{\text{Train length}+\text{Platform length}} {\text{Train speed}}$$

• Example: A 200 m train travels at 72 km/h. It crosses a 300 m platform. Find the time taken. Solution:

$$72\times\frac5{18}=20\text{ m/s}$$

Total distance:

$$200+300=500\text{ m}$$ $$T=\frac{500}{20} =\boxed{25\text{ seconds}}$$

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7. Two Trains Crossing — Opposite Directions

• Formula:

$$\text{Time}= \frac{L_1+L_2}{v_1+v_2}$$

Convert speeds to the same unit before calculating.

• Example: Two trains of lengths 150 m and 200 m run at 60 km/h and 90 km/h towards each other. Find the crossing time. Solution:

$$v_{\text{relative}}=150\text{ km/h} =150\times\frac5{18} =\frac{125}{3}\text{ m/s}$$

Total distance:

$$150+200=350\text{ m}$$ $$T=\frac{350}{125/3} =\boxed{8.4\text{ seconds}}$$

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8. Two Trains Crossing — Same Direction

• Formula:

$$\text{Time}= \frac{L_1+L_2}{|v_1-v_2|}$$

• Example: Two trains of lengths 120 m and 180 m travel in the same direction at 72 km/h and 54 km/h. Find the overtaking time. Solution: Relative speed:

$$72-54=18\text{ km/h}=5\text{ m/s}$$

Total distance:

$$120+180=300\text{ m}$$ $$T=\frac{300}{5} =\boxed{60\text{ seconds}}$$

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9. Train Passing a Moving Person

• Formula: Opposite direction:

$$v_{\text{relative}}=v_{\text{train}}+v_{\text{person}}$$

Same direction:

$$v_{\text{relative}}=|v_{\text{train}}-v_{\text{person}}|$$ $$T=\frac{\text{Train length}}{v_{\text{relative}}}$$

• Example: A 150 m train moves at 54 km/h. A person runs in the opposite direction at 18 km/h. Find the crossing time. Solution:

$$v_{\text{relative}}=54+18=72\text{ km/h} =20\text{ m/s}$$ $$T=\frac{150}{20} =\boxed{7.5\text{ seconds}}$$

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10. Circular Track — Meeting in the Same Direction

• Formula:

$$T=\frac{\text{Track length}}{|v_1-v_2|}$$

for the first meeting after starting together.

• Example: Two runners on a 400 m circular track run at 8 m/s and 12 m/s in the same direction. Find their first meeting time. Solution:

$$v_{\text{relative}}=12-8=4\text{ m/s}$$ $$T=\frac{400}{4} =\boxed{100\text{ seconds}}$$

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11. Circular Track — Opposite Directions

• Formula:

$$T=\frac{\text{Track length}}{v_1+v_2}$$

• Example: Two runners start together on a 500 m track and run in opposite directions at 6 m/s and 4 m/s. Find their first meeting time. Solution:

$$v_{\text{relative}}=6+4=10\text{ m/s}$$ $$T=\frac{500}{10} =\boxed{50\text{ seconds}}$$

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12. Circular Track — Meeting at Starting Point

• Formula: Individual lap times:

$$T_1=\frac Lv_1,\qquad T_2=\frac Lv_2$$

They meet again at the starting point after:

$$\operatorname{LCM}(T_1,T_2)$$

when the lap times are commensurable.

• Example: Two runners complete a 600 m track in 60 s and 90 s respectively. When will they next meet at the starting point? Solution:

$$\operatorname{LCM}(60,90)=180$$

Therefore:

$$\boxed{180\text{ seconds}}$$

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13. Late / Early Arrival

• Formula: If travelling at (v_1) causes (t_1) hours of lateness and at (v_2) causes (t_2) hours of earliness:

$$\frac{D}{v_1}-\frac{D}{v_2}=t_1+t_2$$

Hence:

$$D= \frac{v_1v_2(t_1+t_2)}{v_2-v_1}$$

for (v_2>v_1).

• Example: Walking at 4 km/h, a student is 5 minutes late. At 5 km/h, the student is 4 minutes early. Find the distance. Solution:

$$D=\frac{4\times5\left(\frac5{60}+\frac4{60}\right)} {5-4}$$ $$=20\times\frac9{60} =\boxed{3\text{ km}}$$

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14. Finding Normal Travel Time from Late/Early Data

• Formula: If at speed (v_1) one is (a) hours late and at (v_2) one is (b) hours early:

$$v_1(t+a)=v_2(t-b)$$

where (t) is the scheduled travel time.

• Example: At 4 km/h, a student is 5 minutes late. At 5 km/h, the student is 4 minutes early. Find the scheduled travel time. Solution:

$$4\left(t+\frac5{60}\right) = 5\left(t-\frac4{60}\right)$$ $$4t+\frac13=5t-\frac13$$ $$t=\frac23\text{ hour}$$ $$\boxed{40\text{ minutes}}$$

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15. Average Speed for Equal Distances

• Formula: For equal distances travelled at speeds (u) and (v):

$$\text{Average speed}= \frac{2uv}{u+v}$$

• Example: A car travels equal distances at 40 km/h and 60 km/h. Find its average speed. Solution:

$$\frac{2(40)(60)}{40+60} =\frac{4800}{100} =\boxed{48\text{ km/h}}$$

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16. Average Speed for Equal Time

• Formula: For equal time intervals:

$$\text{Average speed}=\frac{u+v}{2}$$

• Example: A car travels for 2 hours at 40 km/h and 2 hours at 60 km/h. Find the average speed. Solution:

$$\frac{40+60}{2} =\boxed{50\text{ km/h}}$$

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17. Races — Winning by Distance

• Formula: If A beats B by (x) metres in a race of (D) metres:

$$\frac{v_A}{v_B}=\frac{D}{D-x}$$

• Example: In a 100 m race, A beats B by 10 m. If A runs at 10 m/s, find B’s speed. Solution: A’s time:

$$\frac{100}{10}=10\text{ s}$$

B covers 90 m in 10 s:

$$v_B=\frac{90}{10} =\boxed{9\text{ m/s}}$$

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18. Races — Head Start

• Formula: If B receives a head start of (x) metres:

$$\text{Distance covered by A}=\text{Race length}$$ $$\text{Distance covered by B}=\text{Race length}-x$$

Therefore:

$$\frac{v_A}{v_B} =\frac{D}{D-x}$$

• Example: In a 200 m race, A gives B a 40 m head start and both finish together. If A runs at 10 m/s, find B’s speed. Solution: B covers:

$$200-40=160\text{ m}$$

A’s time:

$$\frac{200}{10}=20\text{ s}$$

Therefore:

$$v_B=\frac{160}{20} =\boxed{8\text{ m/s}}$$

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19. Races — Winning by Time

• Formula: If A finishes (t) seconds before B:

$$T_B=T_A+t$$

and:

$$v=\frac DT$$

• Example: A completes a 200 m race in 20 s and beats B by 5 s. Find B’s speed. Solution:

$$T_B=20+5=25\text{ s}$$ $$v_B=\frac{200}{25} =\boxed{8\text{ m/s}}$$

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20. Speed-Time-Distance Ratio

• Formula: For the same distance:

$$v_1:v_2=t_2:t_1$$

For the same time:

$$d_1:d_2=v_1:v_2$$

• Example: A and B cover the same distance in 5 hours and 8 hours respectively. Find their speed ratio. Solution:

$$v_A:v_B=8:5$$

Therefore:

$$\boxed{8:5}$$

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Advanced Variants

21. Circular Track — Number of Meetings in a Given Time

• Formula: For runners in the same direction:

$$\text{Meetings} \approx \frac{\text{Relative distance covered}}{L}$$

where (L) is the track length. For opposite directions, replace relative speed by (v_1+v_2).

• Example: Two runners move on a 400 m track in the same direction at 10 m/s and 6 m/s. How many times will the faster runner catch the slower one in 1,000 seconds, excluding the starting instant? Solution: Relative speed:

$$10-6=4\text{ m/s}$$

Relative distance:

$$4\times1000=4000\text{ m}$$

Number of complete laps:

$$\frac{4000}{400}=10$$

Therefore:

$$\boxed{10\text{ meetings}}$$

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22. Boats/Moving Objects — Relative Speed

• Formula:

$$v_{\text{downstream}}=v_b+v_s$$ $$v_{\text{upstream}}=v_b-v_s$$

where (v_b) is boat speed in still water and (v_s) is stream speed.

• Example: A boat moves at 12 km/h in still water and the stream flows at 3 km/h. Find its downstream speed. Solution:

$$12+3=\boxed{15\text{ km/h}}$$

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23. Pool / Bounce / Back-and-Forth Motion

• Formula: For motion between two fixed boundaries, use total path length:

$$\text{Distance}=2L\times(\text{number of complete traversals})$$

and:

$$\text{Time}=\frac{\text{Total distance}}{\text{Speed}}$$

• Example: A swimmer crosses a 50 m pool and returns to the starting point. If the speed is 2 m/s, find the time taken. Solution: Total distance:

$$50+50=100\text{ m}$$ $$T=\frac{100}{2} =\boxed{50\text{ seconds}}$$

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24. Relative Speed with Changing Speeds

• Formula: Divide the journey into phases:

$$D=D_1+D_2+\cdots$$ $$T=T_1+T_2+\cdots$$

Apply relative speed separately whenever direction or speed changes.

• Example: A travels towards B at 40 km/h while B travels towards A at 60 km/h for 2 hours. Find the distance covered together before meeting. Solution: Relative speed:

$$40+60=100\text{ km/h}$$

Distance:

$$100\times2 =\boxed{200\text{ km}}$$