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Generalization of straight line motion under constant acceleration

Physics Asked by Sohaib Ali Alburihy on July 29, 2021

My question is that, we all know the three equations of straight line motion under constant acceleration,

begin{align}
x & =x_{rm o}+v_{rm o},t+tfrac12 mathrm a,t^2
tag{1d-a}label{1d-a}
v & =v_{rm o}+mathrm a,t
tag{1d-b}label{1d-b}
v^2 & =v_{rm o}^2+2,mathrm aleft(x-x_{rm o}right)
tag{1d-c}label{1d-c}
end{align}

Is my generalization correct?
begin{align}
mathbf r & =mathbf r_{rm o}+boldsymbol v_{rm o},t+tfrac12mathbf a,t^2 quad text{(no difference with that)}
tag{3d-a}label{3d-a}
boldsymbol v & =boldsymbol v_{rm o}+mathbf a,t
tag{3d-b}label{3d-b}
vertboldsymbol vvert^2 & =vertboldsymbol v_{rm o}vert^2+2,mathbf aboldsymbol cdotleft(mathbf r-mathbf r_{rm o}right)
tag{3d-c}label{3d-c}
end{align}

Please explain the general principle of generalization of 1-dimension formulas into 3-dimensions.

And I must add I am very sorry that I print this question not using LaTeX, I really know nothing about it, so I printed it like that, hopefully you will be patient about me.

2 Answers

Though the above generalisations are correct, what we do in most cases is that we resolute (or break) any given motion along mutually perpendicular axes (namely x, y and z axes) and then apply these formula separately along each of the axes as: $$v_x=v_{0x}+a_{0x}t$$ $$x=v_{0x}t+frac{1}{2}a_{0x}t^2$$ $$v_x^2=v_{0x}^2+2a_{0x}x$$ along x-axes. Similarly, $$v_y=v_{0y}+a_{0y}t$$ $$y=v_{0y}t+frac{1}{2}a_{0y}t^2$$ $$v_y^2=v_{0y}^2+2a_{0y}y$$ and $$v_z=v_{0z}+a_{0z}t$$ $$z=v_{0z}t+frac{1}{2}a_{0z}t^2$$ $$v_z^2=v_{0z}^2+2a_{0z}z$$ along y and z axes respectively. The reason why we do is because motion along mutually perpendicular axes are independent of each other and hence we can apply these formula separately along the axes.

Hope it helps.

Correct answer by Deep Bhowmik on July 29, 2021

enter image description here

As shown in Figure-01

begin{align} mathbf r & =mathbf r_{rm o}+boldsymbol v_{rm o},t+tfrac12mathbf a,t^2 tag{01a}label{01a} boldsymbol v & =boldsymbol v_{rm o}+mathbf a,t tag{01b}label{01b} vertboldsymbol vvert^2 & =vertboldsymbol v_{rm o}vert^2+2mathbf aboldsymbol cdotleft(mathbf r-mathbf r_{rm o}right) tag{01c}label{01c} end{align}

Answered by Frobenius on July 29, 2021

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