A train traveling at 90 km/hr is brought to rest by the application of brakes in a distance of 75 m . Find tha time in which train is brought to rest . Also calculate the distance covered by the train in the first half and second half of this time interval
Answers
Explanation:
similarly using those formula we can solve such type of problem
Answer:
➪ Time taken by the train to stop
= 6 seconds
➪ distance travelled in 1st half time
= 56.25 meter
➪ distance travelled in 2nd half time
= 18.25 meter
Step by step explanations :
given that,
A train traveling at 90 km/hr
and it bought to rest
by the application of brakes in a distance of 75 m
here,
we have,
the initial velocity of the train = 90 km/h
= 90 × 1000 m/3600 s
= 25 m/s
distance covered by the train after applying brakes = 75 m
let the time taken by the train to stop be
t
here,
initial velocity(u) = 25 m/s
final velocity(v) = 0 m/s
time taken(t) = t
distance travelled(s) =75 m
by the equation of motion,
v² = u² + 2as
where,
a is the retardation produced by the brakes
putting the values,
(0)² = 25² + 2(a)(75)
0 = 625 + 150a
150a = -625
a = -625/150
a = -25/6
now,
v = u + at
0 = 25 + (-25/6) × t
-25t/6 = -25
t = -25 × 6/-25
t = 6 seconds
so,
Time taken by the train to stop
= 6 seconds
now,
to find,
distance travelled in 1st half time and 2nd half time
here,
time(t) = 3 seconds
acceleration(a) = -25/6 m/s²
initial velocity(u) = 25 m/s
S = ut + ½ at²
S = 25(3) + ½ (-25/6)(3)(3)
S = 75 - 75/4
S = (300 - 75)/4
S = 225/4
S = 56.25
distance travelled in 1st half time
= 56.25 meter
distance travelled in 2nd half time
= 75 - 56.25
= 18.75 m
so,
_________________
➪ Time taken by the train to stop
= 6 seconds
➪ distance travelled in 1st half time
= 56.25 meter
➪ distance travelled in 2nd half time
= 18.25 m