Question

a car traveling at 27 m/s slams of its brakes to come to a stop. it decelerates at a rate of 8 m/s2. what is the stopping distance of the car.

Answer 1

Answer:

28.6m

Explanation:

Here, initial velocity, u=22.4m/s and final velocity, v=0 m/s  

Time taken by car to stop  t=2.55s

If a be the acceleration.

Using formula v=u+at,

0=22.4+a(2.55)

We get  a=−8.78m/s  

2

   (negative signs means deceleration of car)

If d be the distance traveled by car.

Using formula v  

2

−u  

2

=2ad

0  

2

−(22.4)  

2

=2(−8.78)d

⟹ d=28.6m