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

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:

& \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{aligned}$$ Now, integrate both the sides of equation (4), we get: $$\begin{aligned} & \Rightarrow \int{dv}=\int{adt} \\\ & \Rightarrow v=at+c......(5) \\\ \end{aligned}$$ Here $$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{aligned} & \Rightarrow \dfrac{ds}{dt}=at+u \\\ & \Rightarrow ds=atdt+udt......(6) \\\ \end{aligned}$$ Now, integrate both sides of equation (6), we get: $$\begin{aligned} & \Rightarrow \int{d}s=\int{atdt}+\int{udt} \\\ & \Rightarrow s=\dfrac{a{{t}^{2}}}{2}+ut+C......(7) \\\ \end{aligned}$$ Here C is constant of integration which is distance at $$t=0$$ . Since, the distance at $$t=0$$ is zero. Thus, C = 0. Hence, the equation becomes $$\begin{aligned} & \Rightarrow s=\dfrac{a{{t}^{2}}}{2}+ut+0 \\\ & \Rightarrow s=ut+\dfrac{1}{2}a{{t}^{2}} \\\ \end{aligned}$$ **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.