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Itachi, perhaps your calculations should be corrected as follows:

(From the 1st equation)$$y=ut-uk+\frac{1}{2}at^2+\frac{1}{2}ak^2-atk$$ $$\frac{\text{d}y}{\text{d}t}=u+at-ak$$

 

$$y''=a$$

 

(From the 2nd equation)$$y=ut+\frac{1}{2}at^2$$ $$\frac{\text{d}y}{\text{d}t}=u+at$$

 

$$y''=a$$

Itachi, perhaps your calculations should be corrected as follows:

(From the 1st equation)$$y=ut-uk+\frac{1}{2}at^2+\frac{1}{2}ak^2-atk$$ $$\frac{\text{d}y}{\text{d}t}=u+at-ak$$

 

$$y''=a$$

 

(From the 2nd equation)$$y=ut+\frac{1}{2}at^2$$ $$\frac{\text{d}y}{\text{d}t}=u+at$$

 

$$y''=a$$

Itachi, perhaps your calculations should be corrected as follows:

(From the 1st equation)$$y=ut-uk+\frac{1}{2}at^2+\frac{1}{2}ak^2-atk$$ $$\frac{\text{d}y}{\text{d}t}=u+at-ak$$

$$y''=a$$

(From the 2nd equation)$$y=ut+\frac{1}{2}at^2$$ $$\frac{\text{d}y}{\text{d}t}=u+at$$

$$y''=a$$

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Itachi, perhaps your calculations should be corrected as follows:

(From the 1st equation)$$y=ut-uk+\frac{1}{2}at^2+\frac{1}{2}ak^2-atk$$ $$\frac{\text{d}y}{\text{d}t}=u+at-ak$$

$$y''=a$$

(From the 2nd equation)$$y=ut+\frac{1}{2}at^2$$ $$\frac{\text{d}y}{\text{d}t}=u+at$$

$$y''=a$$