A train is travelling at a speed of 90 km h-1 . Beakers are Applied so as to produce a uniform acceleration of 0.5 ms-2, find how far the train will go before it is brought to rest?
Answers
Explanation:
We know, 1\ km = 1000\ m ; 1\ hr = 3600\ s)
Given, Initial speed of the train,u = 90\ kmh^{-1} = \frac{90\times1000}{3600} = 25\ ms^{-1}
Acceleration of the train,a = -0.5\ ms^{-2} (Negative sign implies retardation)
Since, the train has to be brought to rest, final speed of the train, v = 0\ ms^{-1}
We know, v^2 = u^2 + 2as
\\ \implies 0^2 = 25^2 + 2(-0.5)s \\ \implies 0 = 625 -s \\ \implies s = 625\ m
Therefore, the train travels a distance of 625\ m before coming to rest.
Answer:
Correct option is
A
625 m
Given that,
Acceleration a=−0.5m/s2
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=0.525
t=50 sec
Again, using equation of motion,
S=ut+21at2
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=25×50−21×0.5×(50)2
s=625 m
So, the train will go before it is brought to rest is 625 m.
Hence, A is correct.