Question

The brake applied to a running train produce an acceleration of 0.6 msquare in the direction opposite of that to if train takes 1min. Calculate the distance traveled by it in this time

Answer 1

Answer:

1080 metres

Explanation:

since the acceleration is constant we can simply apply equation of motions

v=u-gt

s= ut -1/2 gt^2

from first equation we can find u= 36m/s

now from second equation

s= 1080metres

Answer 2

$\large \bf \red{ GIVEN :} \\ \Longrightarrow \bf a = 0.6 \: {ms}^{ - 2} \\ \Longrightarrow \bf t = 1 \: min \: = 60 \: sec \\ \Longrightarrow \bf v = 0 \: m {s}^{ - 1}$

$\large \bf \red{ TO \: FIND :} \\\Longrightarrow \bf Distance \: travelled \: by \: train \:$

$\large \bf \red{ SOLUTION :} \\ \bf From \:first \:law \: of \:motion \\ </p><p> \boxed{\bf \orange{v = u + at}} \\ \\\Longrightarrow \bf u = v - at \\ \\\Longrightarrow \bf u = 0 - ( - 0.6 \times 60) \\ \\ \Longrightarrow \bf u = 36m {s}^{ - 1} \\ \bf From \:second \:law \: of \:motion\\ \boxed{ \bf \orange{ s = ut + \frac{1}{2} a {t}^{2} }} \\ \\ \Longrightarrow \bf s = 36 \times 60 + \frac{1}{2} \times( - 0.6) \times 3600 \\ \\ \Longrightarrow \bf s = 2160 - 1080 \\ \\ \large \boxed{ \bf \green{ s = 1080 \: m}}$