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Question: A train passes an electric post in 10 second and a bridge of length 2 km in 110 second. The speed of...

A train passes an electric post in 10 second and a bridge of length 2 km in 110 second. The speed of engine is:
(a) 18 kmph
(b) 36 kmph
(c) 72 kmph
(d) 90 kmph

Explanation

Solution

The above problem can be solved by using the formula of the speed. First calculate the length of the train to find the speed of the engine and then find the speed of the engine.

Complete step by step answer
Given: The time for crossing the post is t=10  st = 10\;{\text{s}}, the time for crossing the bridge is T=110  sT = 110\;{\text{s}}, the length of the bridge is L=2  km=2  km×1000  m1  km=2000  mL = 2\;{\text{km}} = 2\;{\text{km}} \times \dfrac{{1000\;{\text{m}}}}{{1\;{\text{km}}}} = 2000\;{\text{m}}.
The equation to find the speed of the engine by using the time to cross the post is given as:
\Rightarrow v1=lt......(1){v_1} = \dfrac{l}{t}......\left( 1 \right)
The equation to find the speed of the engine by using the time to cross the bridge is given as:
\Rightarrow v2=L+lT......(2){v_2} = \dfrac{{L + l}}{T}......\left( 2 \right)
The speed of the engine remains the same to cross the post and bridge, so equate the equation (10 and equation (2) to calculate the length of the train.
\Rightarrow v1=v2{v_1} = {v_2}
lt=L+lT......(3)\dfrac{l}{t} = \dfrac{{L + l}}{T}......\left( 3 \right)
Substitute 2 km for L, 10 s for t and 110 s for T in the equation (3) to find the length of the train.
\Rightarrow l10  s=2  km+l110  s\dfrac{l}{{10\;{\text{s}}}} = \dfrac{{2\;{\text{km}} + l}}{{110\;{\text{s}}}}
11l=2  km+l11l = 2\;{\text{km}} + l
\Rightarrow 10l=2  km10l = 2\;{\text{km}}
l=0.2  kml = 0.2\;{\text{km}}
Substitute 0.2  km0.2\;{\text{km}} for l and 10 s for t in the equation (1) to find the speed of the engine.
\Rightarrow v1=0.2  km10  s{v_1} = \dfrac{{0.2\;{\text{km}}}}{{10\;{\text{s}}}}
v1=0.2  km(10  s×1  h3600  s){v_1} = \dfrac{{0.2\;{\text{km}}}}{{\left( {10\;{\text{s}} \times \dfrac{{1\;{\text{h}}}}{{3600\;{\text{s}}}}} \right)}}
\Rightarrow v1=72  km/h{v_1} = 72\;{\text{km}}/{\text{h}}

Thus, the speed of the engine is 72  km/h72\;{\text{km}}/{\text{h}} and the option (c) is the correct answer.

Note:
The above problem can also be solved by the concept of the relative motion. The resultant speed becomes equal to the sum of the speed of both objects if both objects move in opposite directions and become equal to subtraction of speed of both objects.