B Acceleration -- How is the intermediate displacement speed derived?

AI Thread Summary
The discussion centers on the application of SUVAT equations, particularly in the context of deriving intermediate displacement speed under constant acceleration. It emphasizes the importance of formulating clear questions to effectively solve problems in physics. A proposed formula for intermediate speed is critiqued, with corrections provided to align it with established SUVAT principles. The correct expression for speed is clarified, ensuring it meets the criteria for a valid physical equation. Understanding and applying SUVAT equations is essential for solving related motion problems accurately.
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How is the intermediate displacement speed derived?
I don't know if I wrote it correctly
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Do you know any SUVAT equations?
 
PeroK said:
SUVAT
Thank you very much, I will google it
 
How do you know that the "SUVAT Equations" apply? I don't understand, which problem is to be solved to begin with. Without asking a clear question, you can't get anything done in science!
 
huc369 said:
Summary: How is the intermediate displacement speed derived?

I don't know if I wrote it correctlyView attachment 315157
Since SUVAT involves constant-acceleration,
and constant-velocity (zero acceleration) is a special case,
consider your proposed formula in the zero-acceleration case, where ##v_0=v_B=v_T##
(where B represents the intermediate displacement: ##(x_B-x_0)=\frac{1}{2}(x_T-x_0)## in your time interval ##0\leq t\leq T##).

Assume ##v_B>0##.
Your proposed formula would read
$$v_B=\frac{\sqrt{(v_B)^2+(v_B)^2}}{2}=\frac{v_B}{\sqrt{2}}\qquad\mbox(false).$$
Instead, it should (based on the symbols available in your recollection) be
$$v_B=\sqrt{ \frac{(v_0)^2+(v_T)^2}{2}}$$
so that $$v_B=\sqrt{ \frac{(v_B)^2+(v_B)^2}{2}}=v_B.$$
(Technically, to be a speed, the left-hand side should be ##|v_B|##.)

Now, following @PeroK 's suggestion to your question,
this could be derived using SUVAT (the constant acceleration equations).
 
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