Question

A train running at 108 km/h is brought to a halt in 2 minutes, calculate the retardation produced by the application of breaks. Also calculate the distance the train travels before stopping​

Answer 1

AnSwer :

A train is running at 108 km/h i.e Initial Velocity (u) will be 108 km/hr. The train is brought to rest i.e Final Velocity (v) will be 0 m/s and time Interval (t) to brought the train in rest position is 2 minutes. We are asked to find the retardation produced by the application of breaks of train and also distance the train travels before stopping.

• First of all we have to make the units same. So, we have to convert time interval (t) from minutes to second.

$:\implies\sf Time \: Interval \: (t) = 2 \: minutes$

$:\implies\sf Time \: Interval \: (t) = 2 \times 60\: second$

$:\implies \underline{ \boxed{\sf Time \: Interval \: (t) = 120\: second}}$

• Now, we have convert the initial velocity (u) of the train from km/hr to m/s :

$\dashrightarrow\:\sf Initial \: Velocity (u) = 108 \: Km/hr$

$\dashrightarrow\:\sf Initial \: Velocity (u) = 108 \times \dfrac{5}{18} \: m/s$

$\dashrightarrow\:\sf Initial \: Velocity (u) = \dfrac{540}{18} \: m/s$

$\dashrightarrow\: \underline{ \boxed{\sf Initial \: Velocity (u) = 30 \: m/s }}$

We have make the units same. So, now by using first kinematical equation of motion we can calculate the retardation (negative acceleration) produced by the application of breaks :

$\longrightarrow\:\:\sf v = u + at$

$\longrightarrow\:\:\sf 0= 30 + a \times 120$

$\longrightarrow\:\:\sf - 30 = 120 a$

$\longrightarrow\:\:\sf a = \dfrac{ - 30}{120}$

$\longrightarrow\:\:\sf a = \dfrac{ - 3}{12}$

$\longrightarrow\:\: \underline{ \boxed{\sf a = - 0.25 \: m/s^2}}$

Hence,the retardation produced by the application of breaks is 0.25 m/s².

• Now, by using second kinematical equation of motion we can find the distance the train travels before stopping.

$\leadsto\sf s = ut + \dfrac{1}{2} at^2$

$\leadsto\sf s = 30 \times 120 + \dfrac{1}{2} \times ( - 0.25) \times (120)^2$

$\leadsto\sf s = 3600 + \dfrac{1}{2} \times ( - 0.25) \times 14400$

$\leadsto\sf s = 3600 + ( - 0.25) \times 7200$

$\leadsto\sf s = 3600 + ( - 1800)$

$\leadsto\sf s = 3600 - 1800$

$\leadsto \underline{ \boxed{\sf s =1800 \: m}}$

Hence,the distance the train travels before stopping is 1800 m.