Question

The distance x covered in time t by a body having initial velocity u and having constant acceleration a is given by \[x=ut+\dfrac{1}{2}a{{t}^{2}}\] . This result follows from
a) Newton’s first law
b) Newton’s second law
c) Newton’s third law
d) None of the above

Answer 1

Hint: Newton's second Law of motion states that the rate of change of momentum of a body is equal to the force applied on the body. Also, momentum is defined as the product of mass and velocity of a body.

Complete step by step answer:
From Newton's second Law of motion, we have: \[F=ma......(1)\] ,
where m is mass of the body and a is the acceleration.

Also, we know that: \[F=\dfrac{dp}{dt}......(2)\]
Therefore,
\[\Rightarrow \dfrac{dp}{dt}=ma......(3)\]
Since, \[p=mv\]
So, we have:
\[\begin{align}
  & \Rightarrow \dfrac{d\left( mv \right)}{dt}=ma \\
 & \Rightarrow m\dfrac{dv}{dt}=ma \\
 & \Rightarrow \dfrac{dv}{dt}=a \\
 & \Rightarrow \dfrac{dv}{dt}=a \\
 & \Rightarrow dv=adt......(4) \\
\end{align}\]

Now, integrate both the sides of equation (4), we get:
\[\begin{align}
  & \Rightarrow \int{dv}=\int{adt} \\
 & \Rightarrow v=at+c......(5) \\
\end{align}\]

Here \[c=u\] , which is u initial velocity.
Therefore,
\[v=at+u......(6)\]

Since\[v=\dfrac{ds}{dt}\] , where s is the distance covered by the body.
So, we can write equation (5) as:
\[\begin{align}
  & \Rightarrow \dfrac{ds}{dt}=at+u \\
 & \Rightarrow ds=atdt+udt......(6) \\
\end{align}\]

Now, integrate both sides of equation (6), we get:
\[\begin{align}
  & \Rightarrow \int{d}s=\int{atdt}+\int{udt} \\
 & \Rightarrow s=\dfrac{a{{t}^{2}}}{2}+ut+C......(7) \\
\end{align}\]
Here C is constant of integration which is distance at \[t=0\] .
Since, the distance at \[t=0\] is zero.
Thus, C = 0.
Hence, the equation becomes
\[\begin{align}
  & \Rightarrow s=\dfrac{a{{t}^{2}}}{2}+ut+0 \\
 & \Rightarrow s=ut+\dfrac{1}{2}a{{t}^{2}} \\
\end{align}\]

So, the correct answer is “Option B”.

Note:
While integration always takes constant integration which always gives the initial value of the physical quantity which is getting integrated.Also remember the concept of Newton's second law of motion.