Q. A train running at 108 km/h is brought to a halt in 2 minutes, calculate the retardation produced by the application of breaks. Also calculate the distance the train travels before stopping
Answers
Answer:
Initial speed of train = u = 108 km/h
Final speed of train = v = 0 km/h
Time taken for the change = t = 2 min = 2/60 hr
Retardation produced by breaks =
a = \frac{v - u}{t} = \frac{0 - 108}{ \frac{2}{60} } = - 5400 \: \frac{km}{ {hr}^{2} }a=
t
v−u
=
60
2
0−108
=−5400
hr
2
km
Distance travelled before stopping =
s = ut + \frac{1}{2} a {t}^{2} = 108 \times \frac{2}{60} + \frac{1}{2} \times ( - 5400) \times {( \frac{2}{60} )}^{2} = 3 \: kms=ut+
2
1
at
2
=108×
60
2
+
2
1
×(−5400)×(
60
2
)
2
=3km
Hope this helps.
Answer:
Initial velocity of the train, u = 108 km/s = 30 m/s
It comes to rest. So, its final velocity is 0
Time taken, t = 2 min = 120 s
Retardation produced is given by :
a=\dfrac{v-u}{t}a=
t
v−u
a=\dfrac{0-30\ m/s}{120\ s}a=
120 s
0−30 m/s
a=-0.25\ m/s^2a=−0.25 m/s
2
The retardation produced by the brakes is -0.25\ m/s^2−0.25 m/s
2
.
Now, using third equation of motion :
v^2-u^2=2asv
2
−u
2
=2as
s is the distance travelled
0^2-(30\ m/s)^2=2(-0.25\ m/s^2)s0
2
−(30 m/s)
2
=2(−0.25 m/s
2
)s
s=1800\ ms=1800 m
Step-by-step explanation:
hop it is helpful for you