A train traveling at a uniform speed passes by a platform 220 m long in 30 s and another platform 325 m long in 39 s Find.
(i) the length of train and the speed of train
I need a answer with process if you have no process don't answer otherwise your answer will be reported
Answers
Given s =t+ 220/30 ••••••• (1)
and s= t + 325/39 ••••••• (2)
on solving (1) and (2), we get
39(t + 220) = 30(t + 325)
39t + 8580 = 30t + 9750
9t = 1170
t = 130 metres
Substitute in (1), we get
Speed = 130 + 220/30
= 350/30 =
= 11.66 m/sec
On converting it to km/hr, we get
S = 3600 * 11.66/1000
= 41.9
= ~42
Explanation:
Given , \bold{s = t + \frac{220}{30} ...... eqn(i)}s=t+
30
220
......eqn(i)
\bold{s = t + \frac{325}{39} ......... eqn(ii)}s=t+
39
325
.........eqn(ii)
By solving \bold{eqn(i)}eqn(i) and \bold{eqn(ii)}eqn(ii) , we get
\bold{39 (t + 220)}39(t+220) =\bold{ 30 (t + 325)}30(t+325)
\bold{39t + 8580 = 30t + 9750}39t+8580=30t+9750
\bold{9t = 1170}9t=1170
\bold{t = 130\:m}t=130m
Substituting in \bold{eqn(i)}eqn(i) , we get
Speed = \bold{130 + \frac{220}{30}}130+
30
220
=\bold{ \frac{350}{30}}
30
350
= \bold{11.66\: m / sec}11.66m/sec
On converting in \bold{km/h}km/h
Speed = \bold{\frac {18}{5}\times 11.66\:\:\:km/h}
5
18
×11.66km/h
= \bold{41. 976\:\:\: km/h}41.976km/h