
Two trains \[A\] and $B$ , are running on parallel tracks. One overtake the other in $20s$ and one crosses the other in $10s$ . The velocity of the trains are
$A.$ $5m{s^{ - 1}},5m{s^{ - 1}}$
$B.$ $10m{s^{ - 1}},15m{s^{ - 1}}$
$C.$ $15m{s^{ - 1}},5m{s^{ - 1}}$
$D.$ $15m{s^{ - 1}},30m{s^{ - 1}}$
Answer
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Hint: This problem is based on relative velocity. Relative velocity is the velocity of one object with respect to the other object. If the two objects are moving in the opposite direction the relative velocity is obtained by adding the velocity of two objects $\left( {{V_{AB}} = {V_A} + {V_B}} \right)$ . When two objects are moving in the same direction the relative velocity is obtained by subtracting the velocity of two objects $\left( {{V_{AB}} = {V_A} - {V_B}} \right)$
Complete step-by-step solution:
Let $u$ and $v$ be the velocity of two trains $A$ and $B$ respectively
Total distance $\left( S \right)$ = Length of train $A$ + Length of train $B$
$ S = 100m + 100m \\
S = 200m \\
$
While overtaking, the two trains will travel in the same direction. Therefore relative velocity of train $A$ with respect to train $B$ is
${V_{AB}} = u - v$
$\dfrac{{200}}{{20}} = u - v$
Therefore, $u - v = 10$ ……….. $\left( 1 \right)$
$ \Rightarrow $ While crossing, the two trains will travel in opposite directions. Therefore relative velocity of train $A$ with respect to train $B$ is
${V_{AB}} = u + v$
$\dfrac{{200}}{{10}} = u + v$
Therefore, $u + v = 20$ ………$\left( 2 \right)$
$v = 20 - u$ ………. $\left( 3 \right)$
Substituting equation $\left( 3 \right)$ in equation $\left( 1 \right)$ we get
$ u - \left( {20 - u} \right) = 10 \\
u - 20 + u = 10 \\
2u = 30 \\
u = \dfrac{{30}}{2} \\
$
Therefore, \[u = 15m{s^{ - 1}}\]
Substituting equation $\left( 3 \right)$ in equation $\left( 1 \right)$
$
15 - v = 10 \\
v = 15 - 10 \\
$
Therefore, $v = 5m{s^{ - 1}}$
$\therefore $ Velocity of train $A$ is $15m{s^{ - 1}}$
Velocity of train $A$ is $5m{s^{ - 1}}$
Hence, Option $C$ is correct. That is $15m{s^{ - 1}},5m{s^{ - 1}}$
Note: The difference between relative velocity and velocity is that relative velocity is measured in a frame where an object can be at rest or in motion with respect to the absolute frame. While the velocity is measured with respect to a reference point which is relative to a different point.
Complete step-by-step solution:
Let $u$ and $v$ be the velocity of two trains $A$ and $B$ respectively
Total distance $\left( S \right)$ = Length of train $A$ + Length of train $B$
$ S = 100m + 100m \\
S = 200m \\
$
While overtaking, the two trains will travel in the same direction. Therefore relative velocity of train $A$ with respect to train $B$ is
${V_{AB}} = u - v$
$\dfrac{{200}}{{20}} = u - v$
Therefore, $u - v = 10$ ……….. $\left( 1 \right)$
$ \Rightarrow $ While crossing, the two trains will travel in opposite directions. Therefore relative velocity of train $A$ with respect to train $B$ is
${V_{AB}} = u + v$
$\dfrac{{200}}{{10}} = u + v$
Therefore, $u + v = 20$ ………$\left( 2 \right)$
$v = 20 - u$ ………. $\left( 3 \right)$
Substituting equation $\left( 3 \right)$ in equation $\left( 1 \right)$ we get
$ u - \left( {20 - u} \right) = 10 \\
u - 20 + u = 10 \\
2u = 30 \\
u = \dfrac{{30}}{2} \\
$
Therefore, \[u = 15m{s^{ - 1}}\]
Substituting equation $\left( 3 \right)$ in equation $\left( 1 \right)$
$
15 - v = 10 \\
v = 15 - 10 \\
$
Therefore, $v = 5m{s^{ - 1}}$
$\therefore $ Velocity of train $A$ is $15m{s^{ - 1}}$
Velocity of train $A$ is $5m{s^{ - 1}}$
Hence, Option $C$ is correct. That is $15m{s^{ - 1}},5m{s^{ - 1}}$
Note: The difference between relative velocity and velocity is that relative velocity is measured in a frame where an object can be at rest or in motion with respect to the absolute frame. While the velocity is measured with respect to a reference point which is relative to a different point.
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