Question

A train is travelling at a speed of 90km h-1. brakes are applied so as to produce a uniform acceleration of -0.5 m/s-2 find how far the train will go before it is brought to rest.

Answer 1

as we know.......2as=v²-u²
a= -5m\s square
v=0
u=90km\hr= 25m\s

2(-5)s=(0)²-(25)²
= -10s=-625
s= -625\-10
=62.5 metres

Answer 2

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

★Given :-

a = -0.5 m/s²

u = 90 km/h

★Here we have to convert km/h into m/s :-

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

= 25 m/s

v = 0 m/s

s = ?

${\boxed{\sf\:{Using\;Third\; Equation\;of\;Motion}}}$

v² = u² + 2as

2as = v² - u²

★We can also write :-

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

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

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

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

= 625 m

★Hence we get :-

$\text{Car\;will\;come\;to\;stop\;after\;covering\;625 m}$

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