• Speed = Distance/Time – This shows us how fast or slow an object moves.
• It is defined as the distance travelled divided by the time it took to travel that distance.
• Distance is directly proportional to speed, but time is inversely proportional
• Distance = Speed X Time, and .
• Time = Distance / Speed, -The time taken will reduce as the speed increases, and vice versa. Any simple problem can be solved using these formulas.
  However, when utilizing formulas, it’s also crucial to remember to use the right units.

Each Speed, Distance and Time can be expressed in different units:

• Time: seconds(s), minutes (min), hours (hr)
• Distance: (meters (m), kilometers (km), miles, feet
• Speed: m/s, km/hr

So if Distance = km and Time = hr, then as Speed = Distance/ Time; the units of Speed will be km/ hr.

Now that we’ve established the units of Speed, Time, and Distance, let’s look at the formulas for them.

$Distance = Speed$ x $Time$

$Speed = \frac{Distance\;}{Time\;}$

$Time = \frac{Distance\;}{Speed\;}$

1 km = 1000 m

1 h = 3600 s

So 1 h = $\frac{1000}{3600}$ $\;\; = \frac{5}{18}$

Now $Y\; m/sec\; = (X×\frac{5}{18})$

$1\; m\; = \frac{1}{1000} km$

$1\; sec\; = \frac{1}{3600} h$

$1\; m/sec = \frac{3600}{1000}\;\;$ $= \frac{18}{5}$

Now $Y\; km/h\; = (X×\frac{18}{5})\;km/hr$

An object covers equal distance at speed S1 and other equal distance at speed S2 then his average speed for the distance is   $\frac{2s_{1}s_{2}}{s_{1}+s_{2}}$

$S_{T} = Speed\; of\; Train$

$S_{O} = Speed\; of\; Object$

$L_{T} = Length\; of\; Train$

$L_{O} = Length\; of\; Object$

When Train Crosses a Stationary Object with no Length(e.g. Pole) in time t

$S_{T} = \frac{L_{T}}{t}$

When Train Crosses a Stationary Object with Length LO (e.g. Train Platform) in time t

$S_{T} = \frac{L_{T}+L_{O}}{t}$

When Train Crosses a Moving Object with no Length (e.g. Car has negligible length) in time t

Objects moving in Opposite directions

• $(S_{T}+S_{O}) = \frac{L_{T}}{t}$

Objects moving in Same directions

• $(S_{T}-S_{O}) = \frac{L_{T}}{t}$

When Train Crosses a Moving Object with Length LO (e.g. Another Train treated as an object) in time t

Objects(Trains) moving in Opposite directions

• $(S_{T}+S_{O}) = \frac{L_{T}+L_{O}}{t}$

Objects(Trains) moving in Same directions

• $(S_{T}-S_{O}) = \frac{L_{T}+L_{O}}{t}$

Note – In Case for Train 2 is treated as an object