Question

a train is moving with a velocity of 90 km per hour it is brought to stop by applying brakes which is produced by retardation of 0.5 metre per second square find the velocity of 10 seconds and time taken by the train to come to rest​

Answer 1

${\mathfrak{\purple{\underline{\underline{Answer:-}}}}}$

$\sf{v=20\;m/s}$

$\sf{t=50\;s}$

${\mathfrak{\purple{\underline{\underline{Explanation:-}}}}}$

Given:-

$\sf{u=90\;km/hr = 90\times\dfrac{5}{18} =25\;m/s.}$

$\sf{t=10\;s.}$

$\sf{a=-0.5\;m/s^{2}}$

$\sf{By\;using\;1st\;equation\;of\;motion,}$

$\sf{v=u+at}$

$\sf{v=25-\dfrac{5}{10} \times 10 = 25-5=20\;m/s.}$

${\boxed{\boxed{\bf{v=20\;m/s}}}}$

Now,

$\sf{u=25\;m/s.}$

$\sf{v=0}$

$\sf{a=-0.5\;m/s^{2}}$

$\sf{t=?}$

$\sf{By\;using\;1st\;equation\;of\;motion,}$

$\sf{v=u+at}$

$\sf{0=25-\dfrac{5}{10} \;t}$

${\boxed{\boxed{\bf{t=50\;s}}}}$

Answer 2

Answer:

Answer:−

\sf{v=20\;m/s}v=20m/s

\sf{t=50\;s}t=50s

{\mathfrak{\purple{\underline{\underline{Explanation:-}}}}}

Explanation:−

Given:-

\sf{u=90\;km/hr = 90\times\dfrac{5}{18} =25\;m/s.}u=90km/hr=90×

18

5

=25m/s.

\sf{t=10\;s.}t=10s.

\sf{a=-0.5\;m/s^{2}}a=−0.5m/s

2

\sf{By\;using\;1st\;equation\;of\;motion,}Byusing1stequationofmotion,

\sf{v=u+at}v=u+at

\sf{v=25-\dfrac{5}{10} \times 10 = 25-5=20\;m/s.}v=25−

10

5

×10=25−5=20m/s.

{\boxed{\boxed{\bf{v=20\;m/s}}}}

v=20m/s

Now,

\sf{u=25\;m/s.}u=25m/s.

\sf{v=0}v=0

\sf{a=-0.5\;m/s^{2}}a=−0.5m/s

2

\sf{t=?}t=?

\sf{By\;using\;1st\;equation\;of\;motion,}Byusing1stequationofmotion,

\sf{v=u+at}v=u+at

\sf{0=25-\dfrac{5}{10} \;t}0=25−

10

5

t

{\boxed{\boxed{\bf{t=50\;s}}}}

t=50s