20
If the length of a train and a platform are 500 m and 700 m respectively, find the time taken
by the train moving at 54 km/hr to cross the platform
Answers
Answer:
Speed
=
40
km/hr
=
40
×
5
18
=
100
9
m/s
Time
=
18
seconds
Distance Covered
=
100
9
×
18
=
200
m
Therefore, length of the train
=
200
m
Answer:
Answer:
Given :-
Time taken = 4 sec
Final velocity = 36 km/hr × 5/18 = 10 m/s
Mass = 700 kg
To Find :-
Force applied
Solution :-
Firstly let's find Acceleration by using first equation
As we know that
\huge \fbox{V = u + at}
V = u + at
Here,
V = Final Velocity = 10 m/s
U = Initial velocity = 0 m/s
A = Acceleration = ?
T = Time = 4 sec
\tt \: 10 = 0 + a \times 410=0+a×4
\tt \: 0 - 10 = 4a0−10=4a
\tt \: - 10 = 4a−10=4a
\tt \: a = \dfrac{ - 10}{4} = 2.5a=
4
−10
=2.5
Hence :-
Acceleration of body is 2.5 m/s²
Now,
Let's find Distance travelled by using third Equation
\huge \bf \: {v}^{2} - {u}^{2} = 2asv
2
−u
2
=2as
Here,
V = Final Velocity
U = Initial velocity
A = Acceleration
S = Distance
\tt \: {10}^{2} - {0}^{2} = 2( - 2.5)(s)10
2
−0
2
=2(−2.5)(s)
\tt \: 0 - 100 = - 5s0−100=−5s
\tt \: - 100 = - 5s−100=−5s
\tt \: s \: = \dfrac{ - 100}{ - 5} = 20 \: ms=
−5
−100
=20m
Hence :-
Distance travelled by body is 20 m
Now,
Let's find Force applied
As we know that
\huge \fbox{F = ma}
F = ma
Here,
F = Force Applied
M = Mass
A = Acceleration
\tt \:F = 700 \times 2.5F=700×2.5
\tt \: F = 1750F=1750
Hence :-
Force applied is 1750 N.