Question

A train travelling at a speed of 90 kmh^-1. the brakes are applied so as to produce a uniform acceleration of minus 0.5 m/s² find how far does the train go before it is brought to rest.

Answer 1

Answer:

Explanation:

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s

acceleration , a = -0.5 m/s²

Use formula,

v = u + at

Finally train will be rest so, final velocity, v = 0

0 = 25 - 0.5t

25 = 0.5t ⇒t = 50 sec

Again, use formula,

S = ut + 1/2at²

Where S is distance travelled before stop

S = 25 × 50 - 1/2 × 0.5 × 50²

= 1250 - 1/2 × 0.5 × 2500

= 1250 - 625

= 625 m

Hence, distance travelled = 625m and time taken = 50 sec

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Answer 2

Answer:

Distance travelled (s)=625 m

---------------------------

Given:

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

• Final speed (v) = 0 m/s (Rest)

• Accelration (a) = -0.5 m/s²

Explanation:

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

s → Distance travelled

By substituting values in the equation we get:

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

$hope \: its \: help \: u$