Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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

Last updated date: 11th Aug 2024
Total views: 363.2k
Views today: 5.63k
Verified
363.2k+ views
Hint: 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.

Given: The time for crossing the post is $t = 10\;{\text{s}}$, the time for crossing the bridge is $T = 110\;{\text{s}}$, the length of the bridge is $L = 2\;{\text{km}} = 2\;{\text{km}} \times \dfrac{{1000\;{\text{m}}}}{{1\;{\text{km}}}} = 2000\;{\text{m}}$.
The equation to find the speed of the engine by using the time to cross the post is given as:
$\Rightarrow$ ${v_1} = \dfrac{l}{t}......\left( 1 \right)$
The equation to find the speed of the engine by using the time to cross the bridge is given as:
$\Rightarrow$${v_2} = \dfrac{{L + l}}{T}......\left( 2 \right) The speed of the engine remains the same to cross the post and bridge, so equate the equation (10 and equation (2) to calculate the length of the train. \Rightarrow$${v_1} = {v_2}$
$\dfrac{l}{t} = \dfrac{{L + l}}{T}......\left( 3 \right)$
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.
$\Rightarrow$ $\dfrac{l}{{10\;{\text{s}}}} = \dfrac{{2\;{\text{km}} + l}}{{110\;{\text{s}}}}$
$11l = 2\;{\text{km}} + l$
$\Rightarrow$$10l = 2\;{\text{km}} l = 0.2\;{\text{km}} Substitute 0.2\;{\text{km}} for l and 10 s for t in the equation (1) to find the speed of the engine. \Rightarrow$${v_1} = \dfrac{{0.2\;{\text{km}}}}{{10\;{\text{s}}}}$
${v_1} = \dfrac{{0.2\;{\text{km}}}}{{\left( {10\;{\text{s}} \times \dfrac{{1\;{\text{h}}}}{{3600\;{\text{s}}}}} \right)}}$
$\Rightarrow$${v_1} = 72\;{\text{km}}/{\text{h}}$

Thus, the speed of the engine is $72\;{\text{km}}/{\text{h}}$ 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.