#1
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Help needed solving for t
Hello.
I'm helping out a friend with her physics, and I haven't taken physics in several years and am stuck on something that perhaps someone could assist me with. Solve for t: d=1/2*a*t^2)*(v*t) Gracias! |
#2
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Re: Help needed solving for t
Uh, that equation can't be right.
You mean d = vt + 1/2 at^2 I'm sure, yes? |
#3
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Re: Help needed solving for t
[ QUOTE ]
d=1/2*a*t^2)*(v*t) [/ QUOTE ] Your equation doesnt make sens and has the wrong units. If d is distance, a is acceleration, v is velocit, and t is time than i think the equation u want is d=1/2*a*t^2+v*t where v is the initial velocity. To solve this just subtract both sides by d to make 1/2*a*t^2+v*t-d=0 which can be solved usingt the quadratic formula. |
#4
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Re: Help needed solving for t
yes that's it...that's much better [img]/images/graemlins/smile.gif[/img]
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#5
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Re: Help needed solving for t
Hi there!
Usually when this equation is used to solve for t, the initial velocity of the object is zero, or else the quadratic equation must be used (as pointed out by a previous poster). d = Vt + 1/2at^2 d = (0)t + 1/2at^2 d = 1/2at^2 Solving for t now becomes a heck of a lot easier... 2d = at^2 2d/a = t^2 square root(2d/a) = t I hope this helps! Later... |
#6
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Re: Help needed solving for t
I thought this would help me solve the problem, but maybe someone could give feedback to steer me in the right direction on this one.
A spelunker drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 1.58 s after the stone is dropped. How deep is the hole? |
#7
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Re: Help needed solving for t
1/2 at1^2 = ct2
T = t1 + t2 t2 = T - t1 1/2 at1^2 = cT - ct1 at1^2 + 2ct1 - 2cT = 0 t1 = [-2c +/- sqrt(4c^2 + 8acT) ]/2a t2 = T - t1 = T - [-2c +/- sqrt(4c^2 + 8acT) ]/2a d = ct2 = cT - c/2a [-2c +/- sqrt(4c^2 + 8acT)] We may safely discard the negative root (do you see why?), so d = ct2 = cT - c/2a [-2c + sqrt(4c^2 + 8acT)] = (343 m/s)(1.58s) - (343ms / (2x9.8m/s^2))x[-2x343m/s + sqrt(4x(343m/s^2)^2 + 8x9.8m/s^2 x 343m/s x 1.58s) ] Therefore d = 11.71m. |
#8
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Re: Help needed solving for t
Props. It makes sense, looks correct from what little I know [img]/images/graemlins/smile.gif[/img]
Negative root in this case is not applicable to the problem (wouldn't make sense at all) Just out of curiosity, how long did it take you churn that out? |
#9
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Re: Help needed solving for t
Longer than I would have liked. I tried two ugly ways before I found this (relatively) more elegant solution. Probably 10 minutes (but I have a Ph.D. in physics, so don't feel bad).
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