Help needed solving for t

[Guyon]

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!

[Borodog]

Uh, that equation can't be right.

You mean d = vt + 1/2 at^2 I'm sure, yes?

[mrgold]

[ 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.

[Guyon]

yes that's it...that's much better /images/graemlins/smile.gif

[Mr. Sask]

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...

[Guyon]

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?

[Borodog]

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.

[Guyon]

Props. It makes sense, looks correct from what little I know /images/graemlins/smile.gif

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?

[Borodog]

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).