Question

A train is travelling at the speed of 90km/h brahm are applied so as to produce a Uniform acceleration of -0.5m/s2 . Find How far the train will go before it is brought to rest.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Distance\:travelled=625\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: speed(u) = 90 \: km/h \\ \\ \tt: \implies Acceleration(a) = - 0.5 \: {m/s}^{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Distance \: travel(s) = ?$

• According to given question :

$\tt \circ \: Initial \: speed = 90 \times \frac{5}{18} = 25 \: m/s \\ \\ \tt \circ \: Final \: speed = 0 \: m/s\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {25}^{2} + 2 \times - 0.5 \times s \\ \\ \tt: \implies 0 = 625 - s \\ \\ \tt: \implies - 625 = - s \\ \\ \green{\tt: \implies s = 625 \: m} \\ \\ \green{\tt \therefore Distance \: travelled \: by \: train \: is \: 625 \: m \: after \: applying \: brake}$

$\blue{\huge {\boxed{ \tt Some \: related \: formula}}} \\ \\ \orange{ \tt \circ \: First \: eqn \: of \: motion \to \: v = u + at} \\ \\ \orange{ \tt \circ \: Second\: eqn \: of \: motion \to \: s= u t+ \frac{1}{2} {at}^{2} } \\ \\ \orange{ \tt \circ \: Time \: of \: flight \to \: t= \frac{ 2u \: {sin} \: \theta }{g} } \\ \\ \orange{ \tt \circ \: Range \to \: R = \frac{ {u}^{2} sin \: 2 \theta}{g} } \\ \\ \orange{ \tt \circ \: Maximum \: height \to \: H_{max}= \frac{ {u}^{2} {sin}^{2} \: \theta}{2g} }$

Answer 2

Answer :

625 m

explanation :

plz refer to the attachment above