Alex_Mattucci
  • Alex_Mattucci
If d = p + 0.5at^2, which equation is solved for t?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Alex_Mattucci
  • Alex_Mattucci
@Michele_Laino
Alex_Mattucci
  • Alex_Mattucci
@pooja195
pooja195
  • pooja195
@mathmale

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More answers

just_one_last_goodbye
  • just_one_last_goodbye
@Alex_Mattucci it says "which equation" then there must be answer choices ^_^
Alex_Mattucci
  • Alex_Mattucci
yeah... forgot to put those in
just_one_last_goodbye
  • just_one_last_goodbye
First of, If my memory is not failing me (all that are watching please correct)... we subtract P from both sides
just_one_last_goodbye
  • just_one_last_goodbye
Second we would divide the result expression by 0.5a
just_one_last_goodbye
  • just_one_last_goodbye
Then finally we figure out the square root of both sides ^_^
Alex_Mattucci
  • Alex_Mattucci
A. Sqrt d-p-0.5/a B. Sqrt d+p-0.5/a C. Sqrt d+p/0.5a D. Sqrt d-p/0.5a
just_one_last_goodbye
  • just_one_last_goodbye
Hope that helps c:
Alex_Mattucci
  • Alex_Mattucci
Oh so that would be D!
Alex_Mattucci
  • Alex_Mattucci
A. t=Sqrt d-p-0.5/a B. t=Sqrt d+p-0.5/a C. t=Sqrt d+p/0.5a D. t=Sqrt d-p/0.5a
Michele_Laino
  • Michele_Laino
first step: if I subtract \(p\) from both sides, I get: \[d - p = p + \frac{1}{2}a{t^2} - p\] please simplify
Alex_Mattucci
  • Alex_Mattucci
forgot the t= in those answers above
Alex_Mattucci
  • Alex_Mattucci
that would be d-p=1/2at^2
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
second step: if I multiply both sides by \(2/a\), I get: \[\left( {d - p} \right)\frac{2}{a} = \frac{2}{a} \cdot \frac{a}{2}{t^2}\] please simplify
Alex_Mattucci
  • Alex_Mattucci
d-p/0.5a=t^2
Michele_Laino
  • Michele_Laino
hint: left side is: \[\left( {d - p} \right)\frac{2}{a}\] right side: what is: \[\frac{2}{a} \cdot \frac{a}{2} = ...?\]
Alex_Mattucci
  • Alex_Mattucci
i don't know
Michele_Laino
  • Michele_Laino
hint: |dw:1449944810852:dw|
Alex_Mattucci
  • Alex_Mattucci
oh so they cancel out
Michele_Laino
  • Michele_Laino
yes! so, what do you get?
Alex_Mattucci
  • Alex_Mattucci
t^2
Alex_Mattucci
  • Alex_Mattucci
on the right side
Michele_Laino
  • Michele_Laino
that's right! so we can write this: \[{t^2} = \left( {d - p} \right)\frac{2}{a}\]
Michele_Laino
  • Michele_Laino
finally, I take the square root of both sides, so I can write: \[\huge \sqrt {{t^2}} = \sqrt {\left( {d - p} \right)\frac{2}{a}} \] please simplify the left side
Alex_Mattucci
  • Alex_Mattucci
t
Michele_Laino
  • Michele_Laino
that's right! so we get: \[\huge t = \sqrt {\left( {d - p} \right)\frac{2}{a}} \] more precisely, we have: \[\Large \sqrt {{t^2}} = \pm t\] since time is positive, then we take the positive square root
Alex_Mattucci
  • Alex_Mattucci
ok
Michele_Laino
  • Michele_Laino
:)
Alex_Mattucci
  • Alex_Mattucci
so now what?
Michele_Laino
  • Michele_Laino
we have finished, you got the right answer: \[t = \sqrt {\left( {d - p} \right)\frac{2}{a}} \]
Alex_Mattucci
  • Alex_Mattucci
but those are not a, b, c, or d
Michele_Laino
  • Michele_Laino
we can write the right side as below: \[\Large t = \sqrt {\left( {d - p} \right)\frac{2}{a}} = \sqrt {\frac{{d - p}}{{a/2}}} = \sqrt {\frac{{d - p}}{{0.5a}}} \] since \(1/2=0.5\)
Alex_Mattucci
  • Alex_Mattucci
oh ok, that makes more sense!
Michele_Laino
  • Michele_Laino
:)
Michele_Laino
  • Michele_Laino
so, what is the right option?
Alex_Mattucci
  • Alex_Mattucci
Thank you so much for all your help! The answer is D!
Michele_Laino
  • Michele_Laino
that's right!

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