Question

A man walking at the rate of 4 km/hr crosses a bridge in 18 min. Find the length of the bridge in meters. \[\]

Answer 1

Hint: We convert the given speed $v$ of the man from km/hr to m/s by multiplying $\dfrac{5}{18}$ to the given speed 4 km/hr. We convert time taken $t$ from minutes to seconds by multiplying 60 . We find the distance covered by the man in metres as $s=v\text{ m/sec }\times \text{t sec}$. \[\]

Complete step by step answer:
We know that the distance covered by an object is the amount of displacement from starting point to finishing point without considering direction. It is conventionally denoted as $s$. If the time takes taken to cover the distance is $t$ then the speed of the object denoted as $v$ is given by the formula
\[\begin{align}
  & v=\dfrac{s}{t} \\
 & \Rightarrow s=v\times t \\
\end{align}\]
The distance is conventionally measured in meters, kilometres etc, the time is measured in seconds , minutes or hours and the speed is measured in meter per second m/s, kilometre per hour km/hr etc. \[\]
We are given the question that a man walking at the rate of 4 km/hr crosses a bridge in 18 minutes. \[\]
We know that 1 km is equal to 1000mtres and 1 hour is $60\times 60=3600$ seconds. So we have
\[1\text{km/hr}=\dfrac{1000\text{m}}{3600\text{s}}=\dfrac{5}{18}\text{m/s}\]
So we convert and find the speed in m/s as
\[v=4\text{km/hr}=4\times \dfrac{5}{18}\text{m/s}=\dfrac{20}{18}\text{m/s}=\dfrac{10}{9}\text{m/s}\]
We convert the given time take into seconds and have;
\[t=18\text{ minutes}=18\times 60\text{ s}=1080\text{ s}\]
So the length bridge is equal to the distance covered by the man in 1080 sec and with $\dfrac{10}{9}$m/s speed which is
\[s=v\times t=\dfrac{10}{9}\times 1080=1200\text{ m}\]

Note: We can alternatively solve by first converting the time taken in hours $\dfrac{18}{60}$and the finding distance in km $\left( 4\times \dfrac{18}{60}=1.2 \right)$ which we then convert the distance to metres$\left( 1.2\times 1000=1200 \right)$. We note that we have assumed the length of the man as 0. If a bus or truck has to cross we have to exclude the length of the bus and truck.