Question

Two trains 128 m and 122 m long are running towards each other on parallel tracks at 48 km/h and 42 km/h respectively. In what time will they clear off each other from the moment they meet?

Answer 1

Answer:

speed48+42=90km/h

90000/3600m/s

25m/s

distance 128+122=150m

time =150/25=6s

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Train "A" Length = 128 meters.
• Train " B" Length = 122 meters.
• Speed of Train "A" be = 48 Km/h.
• Speed of train "B" be = 42 Km/h.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

As they will clear off,

I.e They should be fully passed ,

They are moving in opposite directions,

So, the relative displacement that should be covered by them is:-

$\large{\bold{ \tt S_r = S_A + S_B}}$

Substituting the values,

$\large{ \tt S_r = 128 + 122}$

Now,

$\large{\boxed{ \tt S_r = 250 \: meters}}$

Here :-

• ${\tt S_A \: represents \: Length \: of \: Train "A"}$
• ${\tt S_B \: represents \: Length \: of \: Train "B"}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, The relative velocity of the train,

$\large{\bold{ \tt V_r = V_A + V_B}}$

Substituting the values,

$\large{ \tt V_r = 48 + 42}$

$\large{ \tt V_r = 90 \: Km/h}$

But we need to convert km/h into m/s.

Now,

$\large{ \tt V_r = 90 \times \dfrac{5}{18}}$

$\large{ \tt V_r = \cancel{90} \times \dfrac{5}{ \cancel{18}}}$

$\large{ \tt V_r = 5 \times 5}$

$\large{\boxed{ \tt V_r = 25 \: m/s}}$

Here:-

• ${\tt V_A represents \: Speed \: of \: Train "A"}$
• ${\tt V_B represents \: Speed \: of \: Train "B"}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

As we know,

$\large{\bold{ \tt Time \: taken = \dfrac{Displacement}{Speed}}}$

Now,

$\large{ \tt T = \dfrac{S_r}{V_r}}$

Substituting the values,

$\large{ \tt T = \dfrac{250}{25}}$

$\large{ \tt T = \dfrac{ \cancel{250}}{\cancel{25}}}$

$\huge{\boxed{\boxed{\tt T = 10 \: Seconds}}}$

$\rule{300}{1.5}$

Note:-

• Here "T" represents time taken by the trains to cross each other.