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2.Motion in Straight Line
easy
If a train travelling at $72\,\, kmph$ is to be brought to rest in a distance of $200$ metres, then its retardation should be............$m{s^{ - 2}}$
A$20 $
B$10$
C$2$
D$1 $
Solution
(d) $u = 72\;kmph = 20m/s,$ $v = 0$
By using ${v^2} = {u^2} – 2as$
$⇒$ $a = \frac{{{u^2}}}{{2s}}$
$ = \frac{{{{(20)}^2}}}{{2 \times 200}} = 1\;m/{s^2}$
By using ${v^2} = {u^2} – 2as$
$⇒$ $a = \frac{{{u^2}}}{{2s}}$
$ = \frac{{{{(20)}^2}}}{{2 \times 200}} = 1\;m/{s^2}$
Standard 11
Physics
