Question

A train is travening of a speed of
90 kmh Brakes are applied so as to
produce a uniform accleration of
-0.5 ms. find how far the train
will go before it is brought to rest?
​

Answer 1

v = 90kmh = 25m/s now @ = -0.5 now

v = 0

hencev²- u² = 2as

or 625 = s

hence it will go 625 m before being at rest

Answer 2

Answer:

$\boxed{\mathfrak{Distance \ travelled \ (s) = 625 \ m}}$

Distance travelled (s)=625 m

Given:

Initial speed (u) = 90 km/h = 25 m/s

Final speed (v) = 0 m/s (Rest)

Acceleration (a) = -0.5 m/s²

$\begin{gathered}\rm From \: 3^{rd} \: equation \: of \: motion \: we \: have: \\ \boxed{ \bf{ {v}^{2} = {u}^{2} + 2as}}\end{gathered}$

From 3 rd

equation of motion we have:

v 2 =u 2 +2as

s → Distance travelled

By substituting values in the equation we get:

$\begin{gathered}\rm \implies {0}^{2} = {25}^{2} + 2( - 0.5)s \\ \\ \rm \implies 0 = 625 - 1s \\ \\ \rm \implies s = 625 \: m\end{gathered}$

⟹0 2 =25 2 +2(−0.5)s

⟹0=625−1s

⟹s=625m