Question

above is the question​

Answer 1

Step-by-step explanation:

We Have :-

$A \: train \: travelling \: at \: a \: speed \: of \: 90 \: km \: h^{ - 1} \\ \\ \\ Brakes \: are \: applied \: to \: it \: to \: stop \: it \\ \\ \\ Brakes \: produce \: an \: accleration \: of \: - 0.5 \: m \: s ^{ - 1}$

To Find :-

$Distance \: travelled \: before \: stopping$

Formula Used :-

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

Solution :-

$First \: we \: need \: to \: convert \: 90 \: km \: h^{ - 1} \: into \: m \: s^{ - 1} \\ \\ \\ 1 \: km \: h^{ - 1} = \frac{1}{3.6} m \: s^{ - 1} \\ \\ \\ 90 \: km \: h^{ - 1} = \frac{90}{3.6} m \: s^{ - 1} \\ \\ \\ = 25m \: s^{ - 1} \\ \\ \\ Now \: using \: the \: formula \\ \\ \\ v^{2} = u^{2} + 2as \\ \\ \\ (0)^{2} = (25)^{2} + 2( - 0.5)s \\ \\ \\ 0 = 625 - s \\ \\ \\ s = 625 \: m \\ \\ \\ The \: train \: stopped \: after \: travelling \: 625 \: m$

Answer 2

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