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anonymous

  • 5 years ago

An equation in the study of kinematics (the description of motion) is, v^2 = vo^2 + 2ax where x is the distance in meters, v is the speed in m/s, a is the acceleration in m/s^2, and t is the time in seconds. Based on this information, is the following equation also possible: x = (v + 1/2 a) t Why or why not?

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  1. anonymous
    • 5 years ago
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    No. Let's begin with these formulas \[v^2={v_0}^2+\text{2ax....Eq 1}\] \[v = v_0+\text{at}....\text{Eq 2} \rightarrow \text{x} = \frac{v^2-{v_0}^2}{2a}\] substituting Eq 2 for v and expanding we get \[\text{x = }\frac{{v_0}^2+2{v_0}\text{at}+a^2t^2-{v_0}^2}{2a}\Rightarrow \frac{2v_0\text{at}+a^2t^2}{2a}\Rightarrow(\frac{2v_0\text{at}}{2a}+\frac{a^2t^2}{2a})\Rightarrow v_0t+\frac{1}{2}\text{a}t^2\] Futher simplification yields; \[\text{x = }(v_0+\frac{1}{2}\text{at})\text{t } \neq (v_0+\frac{1}{2}\text{a})\text{t}\] So no based on your information the following equation can't be obtained \[\text{x = }(v_0+\frac{1}{2}\text{a})\text{t}\]

  2. anonymous
    • 5 years ago
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    Is there a typo in your question?

  3. anonymous
    • 5 years ago
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    Nope. No typo. My Physics professor is just nuts and doesn't make sense all the time. Therefore, I rely on outside help. So I thank you lol.

  4. anonymous
    • 5 years ago
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    you can tell this is wrong simply by dimensional analysis: Let L = length, T = time. Then you can see that \[x=(v+1/2a)t\] is dimensionally\[L=(L/T + L/T ^{2})T = L/T*T + L/T^{2}*T = L + L/T\]which is dimensionally correct. To be correct, you would need every term to have dimensions of L

  5. anonymous
    • 5 years ago
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    Thank-you very much.

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