1. Relative Speed (Same Direction):
\[
\text{Speed of Train} = 15 \, \text{m/s}
\]
\[
\text{Time to Overtake} = 10 \, \text{s}
\]
\[
\text{Distance} = \text{Length of Train} = 100 \, \text{m}
\]
\[
\text{Relative Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{10} = 10 \, \text{m/s}
\]
\[
\text{Speed of Man} = 15 - 10 = 5 \, \text{m/s}
\]

2. Relative Speed (Opposite Direction):
\[
\text{Relative Speed} = 15 + 5 = 20 \, \text{m/s}
\]

3. Time to Cross Man (Opposite Direction):
\[
\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{100}{20} = 5 \, \text{s}
\]

Thus, the train will take 5 seconds to cross the man if he is running in the opposite direction.

When particle comes back to initial position displacement becomes zero.

When objects move in opposite directions relative velocity= v1+ v2= 20 m/s

When moving in same direction relative velocity = V ₁ -V ₂= 10-20 = -10 m/s

Velocity of A w.r.t B = V _A - V _B
Velocity of B w.r.t A = V _B - V _A

Relative Velocity = V _A - V _B
10= 40 - V _B
V _B = 30 m/s

* Is 50 m/s a possible answer ?

As, 54 km/hr = 15 m/s both have same velocity in same direction their relative speed/velocity is zero

As both A and B have same velocity their relative speed is zero. so both have same slope in position time graph.

Using the formula

$a = \frac{v_f - v_i}{t}$

:

Converting to m/s:

$90$

km/h =

$25$

m/s,

$54$

km/h =

$15$

m/s.

$$a = \frac{15 - 25}{15} = \frac{-10}{15} = -0.67 \text{ m/s}^2$$

Retardation = 0.67 m/s².

Using

$t = \frac{s}{v}$

, with

$v = 54$

km/h = 15 m/s and

$s = 5$

m:

$$t = \frac{5}{15} = 0.33 \text{ s}$$

Reaction time = 0.33 s.