The initial velocity of a train is 48.27km per hour.It travels a distance of 110.642m with uniform retardation when its velocity becomes 24.135km per hour.If the train continues to move with this retardation then how far will it travel before stopping?
Answers
Answer:
The initial velocity of train [u] = 60 km/hr
The final velocity of train [v] = 15 km/hr
The initial distance travelled [s] = 450 m
We have 1 km = 1000 m.
1000 m = 1 km
1\ m=\frac{1}{1000}\ km1 m=
1000
1
km
450\ m=\frac{1}{1000} *450\ km450 m=
1000
1
∗450 km
450\ m=\ 0.45\ km450 m= 0.45 km
Let the acceleration of the particle is a .
From equation of kinematics we know that -
v^2-u^2=2asv
2
−u
2
=2as
15^2-60^2\ =\ 2*a*0.4515
2
−60
2
= 2∗a∗0.45
225-3600=0.90a225−3600=0.90a
\ -3375=0.90a −3375=0.90a
a=\frac{-3375}{0.90}\ km/hr^2a=
0.90
−3375
km/hr
2
a=\ -3750\ km/hr^2a= −3750 km/hr
2
Here, negative sign is due to the fact that it is retardation.
Now, we are asked to to calculate the stopping distance.
For stopping distance, the final velocity [v] = 0
The initial velocity [u] = 15 km/hr
From equation of kinematics we know that-
v^2-u^2=2asv
2
−u
2
=2as
0-15^2=2*[-3750]*s0−15
2
=2∗[−3750]∗s
-225=-7500*s−225=−7500∗s
s=\frac{225}{7500}\ kms=
7500
225
km
=0.03\ km=0.03 km [ans]