Question

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?
Note-1)No Copied and spam Answers
2) Should be solved on paper or should be quality nd correct Answer
3) solve it without graph method..means just solve simply by using 3 equations​

Answer 1

Answer:

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.