A train is travelling at a speed
of 90 km h-! Brakes are applied
so as to produce a uniform
acceleration of - 0.5 m s2. Find
how far the train will go before it
is brought to rest.
Answers
Answer:
Given that,
Acceleration a=−0.5m/s
2
Speed v=90km/h=25m/s
Using equation of motion,
v=u+at
Where,
v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the equation
Finally train will be rest so, final velocity,v=0
0=25−0.5t
25=0.5t
t=
0.5
25
t=50 sec
Again, using equation of motion,
S=ut+
2
1
at
2
Where, s = distance
v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the equation
Where S is distance travelled before stop
s=25×50−
2
1
×0.5×(50)
2
s=625 m
So, the train will go before it is brought to rest is 625 m.
Answer:
Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s
acceleration , a = -0.5 m/s²
Use formula,
v = u + at
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒t = 50 sec
Again, use formula,
S = ut + 1/2at²
Where S is distance travelled before stop
S = 25 × 50 - 1/2 × 0.5 × 50²
= 1250 - 1/2 × 0.5 × 2500
= 1250 - 625
= 625 m
Hence, distance travelled = 625m and time taken = 50 sec