Question

A train is travelling aur a speed of 90km/h. Brakes are applied so as to produced a uniform acceleration of -0.5m/s. Find how far the train will go before it is brought to rest.



Tomorrow is my exam and I want the solution

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★Given :-

a = -0.5m/s²

u = 90 km/hr

★Convert it into m/s

$\tt{\rightarrow\dfrac{90\times 1000m}{60\times 60}=25m/s}$

v = 0 m/s

★To Find :-

s = ?

★Solution :-

$\textbf{\boxed{Third\; Equation\;of\;Motion}}$

${\boxed{v^{2} = u^{2} + 2as}}$

★We can also write :-

${\boxed{s=\dfrac{v^{2}-u^{2}}{2a}}}$

★Substitute the values we get :-

$\tt{\rightarrow a =\dfrac{(0)^{2}-{25}^{2}}{2\times(-0.5)}}$

$\tt{\rightarrow a =\dfrac{-25\times {25}^2}{-2\times 0.5}}$

$\tt{\rightarrow a =\dfrac{625}{1}}$

= 625 m

$\boxed{\begin{minipage}{11 cm} Additional Information \\ \\ \ Distance = Speed\times Time \\ \\ Displacement=Velocity\times time \\ \\ Average\; Speed = \dfrac{Initial\:\:Speed+Final\:\:Speed}{2} \end{minipage}}$

Answer 2

Here's your answer ....

initial velocity (u) =90km/h

= 25m/s

final velocity(v) = 0

accliration(a) = -0.5m/s²

distance (s) = ???

use 2nd formula of motion :

${v}^{2} - {u}^{2} = 2as \: \\ s = \frac{ {v}^{2} - {u}^{2} }{2a} \\ s = \frac{ {0}^{2} - {25}^{2} }{2( - 0.5)} \\ s = \frac{ - 625}{ - 1} \\ s = 625 \: m$

Hence the body will cover distance of 625 m.