A train moving with a velocity of 90km/hr. It is brought to stop by applying the brakes which produce a retardation of 0.5m/s^2.Find:
(i)the velocity after 10 second
(ii)time taken by the train to come to rest
Answers
Heya mate !!
v is final velocity
u is initial velocity
a is retardation ( acceleration ).
We know that ,
v = u -at
hence , velocity at the time of applying break
= 90km/h × (5/18). (we multiply 5/18 to convert
(we multiply 5/18 to convert km/h directly into m/s to get the exact number with surety )
= 25m/s (u)
1. velocity after 10s
u = 25 .( initial velocity )
u = 25 .( initial velocity ) a = 1/2 ( retardation )
u = 25 .( initial velocity ) a = 1/2 ( retardation ) t = 10s( time )
=> V = 25-(1/2) ×10
=> v = 20 m/s.
2.time to come at rest
=> 0 = 25-(1/2)×t
=> t = 50 seconds
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