A train is travelling at a speed of 90 km /hr break are applied in train to produce uniform acceleration 1.5 m/s find distance for train before it comes to rest
Answers
Explanation:
Given :
final velocity (v) = 0 m/sec
Initial velocity (u) = 90 km/hr ( 25 m/sec )
acceleration (a) = -1.5 m/sec
distance (s) = ???
ACCORDING TO THE FORMULA :-
2as = v^2 - u^2
2×-1.5× s = 0^2 - 25^2
-3 × s = 0 - 625
-3s = -625
s = -625/-3
s = +208.33 m
Therefore ,
distance = 208.33 meters
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Answer:
625 m
Explanation:
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.