A car is travelling with a speed of 10m/s. Breaks are applied and the car retards at the rate of 0.5 m/s². In how much time the car will come to rest.
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Answered by
0
Explanation:
Correct option is
A
625 m
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.
Hence, A is correct.
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