B. velocity at position x, velocity at position x=0, position x, and the original position
In the equation
=
+2 a x (x - x₀)
= velocity at position "x"
= velocity at position "x = 0 "
x = final position
= initial position of the object at the start of the motion
Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.
However, photons are considered to have a rest mass zero.
However note the point carefully "rest mass". A body in motion cannot have mass to be zero.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>
Answer:
Explanation:
Given
time taken ![t=2\ s](https://tex.z-dn.net/?f=t%3D2%5C%20s)
Speed acquired in 2 sec ![v=42\ m/s](https://tex.z-dn.net/?f=v%3D42%5C%20m%2Fs)
Here initial velocity is zero ![u=0](https://tex.z-dn.net/?f=u%3D0)
acceleration is the rate of change of velocity in a given time
![a=\frac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7D%7Bt%7D)
![a=\frac{42-0}{2}=21\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B42-0%7D%7B2%7D%3D21%5C%20m%2Fs%5E2)
Distance travel in this time
![s=ut+0.5at^2](https://tex.z-dn.net/?f=s%3Dut%2B0.5at%5E2)
where
s=displacement
u=initial velocity
a=acceleration
t=time
![s=0+\0.5\times 21\times (2)^2](https://tex.z-dn.net/?f=s%3D0%2B%5C0.5%5Ctimes%2021%5Ctimes%20%282%29%5E2)
![s=42\ m](https://tex.z-dn.net/?f=s%3D42%5C%20m)
so Jet Plane travels a distance of 42 m in 2 s