Come here pls..A driver of a car travelling at 52 kilometre per hour applies the brakes and accelerates uniformly in the opposite direction. the car stops in 5 seconds .another driver going at 3 kilometre per hour in another car applies is break slowly and stop in 10 seconds. . which of the car travel father after the brakes are applied?
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Explanation:
For 1st car
u=52kmph,t=5s
Convert the kmph into m/s. To do that,multiply 52 by 5/18.
It gives us 14.44 m/s.
Now, s=ut+1/2at^2
But we dont know whats a
a=v-u/t
v is not given directly but we can find it. Since the object is stopping, its obvious that final velocity is 0.
a=0-14.44/5
a=-2.88 (- shows us that velocity is decreasing)
Now, s= ut+1/2at^2
s= 14.44×5 +1/2×(-2.88)×25
We will get s as 36.2 m.
For 2nd car
3kmph= 0.83 m/s,t=10
a=v-u/t
a=-0.083 m/s^2
s= 0.83×10+1/2×(-0.083)×100
We will get s as 4.15m.
So, car a travels more than car b.
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