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The distance \mathrm{x} covered in time  \mathrm{t } by \mathrm{a } body having initial velocity v_{o} and having a constant acceleration \mathrm{a } is given by \mathrm{x=v_0 t+(1 / 2) a t^2, }  This result follows -

Option: 1

Newton's first law


Option: 2

 Newton's second law 


Option: 3

 Newton's third law 


Option: 4

None of these 


Answers (1)

best_answer

We know that, 
\mathrm{\vec{F}=m \vec{a} \, \, \, \, \, \, \,} \\ {\vec{a}=\frac{d^2 x}{d t^2}}
From the equation given in the question -
\mathrm{ \frac{d x}{d t}=v_0+a t }
\mathrm{\begin{aligned} & \frac{d^2 x}{d t^2}=0+a \\ & \frac{d^2 x}{d t^2}=a=\text { Acceleration. } \end{aligned}}
Thus equation  \mathrm{x=v_0 t+(1 / 2) a t^2, } at follow Newton's second law.

Posted by

shivangi.bhatnagar

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