As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]



Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Visceral epithelial cells
Given that,
Mass of the stone, m = 400 g = 0.4 kg
Initial speed, u = 20 m/s
It is climbed to a height of 12 m.
To find,
The work done by the resistance force.
Solution,
Let v is the final speed. It can be calculated by using the conservation of energy.

Work done is equal to the change in kinetic energy. It can be given as follows :

So, the required work done is 32.99 J.
It would be Atoms, they’re all made up of these tiny particles