Question

A train is travelling at a speed of 75 km/h. The brakes are applied so as to produce a uniform

acceleration of -0.5 m/s2

.Find how far the train goes before it stops?​

Answer 1

Explanation:

u = 75 km/hr = 20.83 m/s

a = - 0.5 m/s^2

v = 0

v^2 - u^2 = 2×a×s

- (20.83 × 20.83) = 2× (- 0.5) × s

s = (20.83)^2 = 434.02 m (approx)

Answer 2

A train travelling at 90km/hr is brought to rest by the application of brakes in a distance of 75m.find the time in which the train is brought to rest Also calculate the distance covered by the train in the first half and the second half of this time interval.





Using equation of motion for calculating the time in which the train is brought to rest:
v
2
=u
2
+2as
⇒a=−54000km/h
2

0
2
=90
2
+2a(75×10
−3
)
v=u+at
0=90+(−54000)t
⇒t=0.00166h=0.00166×60min=0.1min=0.1×60s=6s
So, the train comes to rest in 6 s.
Distance traveled in the first half, i.e., distance traveled in 3s:
v=90+(−54000)(
3600
3
​
)=45km/h
(45)
2
=90
2
+2(−54000)s
s=56.25m
Thus, distance travelled in second half = (75-56.25)m = 18.75 m