A train is moving with a velocity of 90kmh-1. It is bought to stop by applying the brakes which produce a retardation of 0.5 ms-2 . Find :1. The velocity after 10 sec and 2. The time taken by the train to comes at rest
Answers
V= u+at
First change Km/h into m/s
= 90×5/18
= 25m/s
1.
=25+ 0.5×10
= 25+5
=30
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