Here is a shortcut to memorize the formulas for Time, Speed and Distance.

\(Distance = Time \times Speed\)

\(Time = \frac{Distance}{Speed}\)

\(Speed = \frac{Distance}{Time}\)

\(Average \; Speed = \frac{Total \; Distance}{Total \; Time}\)

## Shortcut Tips & Formulas to Solve Time, Speed & Distance Problems Faster

**Km/hr to m/sec Conversion**

**m/sec to km/hr Conversion**

- If the ratio of the speeds of A and B is a : b, then the ratio of the

the time taken by them to cover the same distance is –

\(\frac{1}{a} : \frac{1}{b} \; or \; b:a\)

- If a train covers a certain distance x km/hr and an equal distance at y km/hr. Then,

Average Speed during the whole journey is \(\left ( \frac{2xy}{x + y} \right ) \; km/hr\)

**Example 1 – A train traveling at 108 kmph crosses a platform in 30 seconds and a man standing on the platform in 20 seconds. What is the length of the platform in meters?**

Extra time taken by the train to cross the platform = 30 – 20 = 10 sec

\(Length \; of \; the \; platform = speed \; of \; train \times extra\; time \; taken \; to \; cross \; the \; platform.\) \(Length \; of \; the \; platform = 108 \; km/hr \times 10 \; sec\)\(Length \; of \; the \; platform = 108 \; km/hr \times 10 \; sec\)Convert 108 km/hr into m/sec\(108 \; km/ph = \frac{5}{18} \times 108 = 30 \; m/sec\)

Convert 108 km/hr into m/sec\(108 \; km/ph = \frac{5}{18} \times 108 = 30 \; m/sec\)

\(108 \; km/ph = \frac{5}{18} \times 108 = 30 \; m/sec\)Therefore, length of the platform = \(30 \; m/s \times 10 \; sec = 300 \; meters\)

Therefore, length of the platform = \(30 \; m/s \times 10 \; sec = 300 \; meters\)

**Example 2 – Karen travels the first 4 hours of his journey at 75 mph speed and the remaining 5 hours at 35 mph speed. What is the average speed of Jim’s travel in mph?**

Total distance travelled and total time taken –

\(Average \; Speed = \frac{Total \; Distance}{Total \; Time}\)Total distance traveled by Karen = Distance covered in the first 4 hours + Distance covered in the next 5 hours.

Distance covered in the first 4 hours \(= 4 \times 75 = 300 \; miles\)

Distance covered in the first 5 hours \(= 5 \times 35 = 175 \; miles\)

Therefore, total distance traveled = 300 + 175 = 475 miles.

Total time taken = 4 + 5 = 9 hours.

Therefore,

\(Average \; Speed = \frac{Total \; Distance}{Total \; Time}\) \(= \frac{475}{9}\) \(= 52.77 \; m/h\)**Also Read:**

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