Rearranging Formula with t as Subject - Grace

[physicsgirl_nz]

Hey guys,

I'm new here and I need help on rearranging a formula so that t is the subject.

d = vit+1/2at*2

How do you use LAtex software?

thanks

Grace

 

[cookiemonster]

A thread on LaTeX as it applies to physicsforums.com.

https://www.physicsforums.com/showthread.php?s=&threadid=8997

After that, I just read through this little thing,

http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf

and I was pretty much set.

That being said, onto your question. I assume you meant:

d = v_it + \frac{1}{2}at^2.

This is a quadratic equation in the variable t. Rewrite it this way:

\frac{1}{2}at^2 + v_it - d = 0
which corresponds to, when compared to the general quadratic equation of the form
\alpha x^2 + \beta x + \gamma = 0

\alpha = \frac{1}{2}a
\beta = v_i
\gamma = -d
x = t

which can be solved with:

x = \frac{-\beta \pm \sqrt{\beta^2 - 4\alpha\gamma}}{2\alpha}

It was a good exercise in LaTeX! Now you try.

cookiemonster

 

[physicsgirl_nz]

Okay, the problem involving the formula was a ball bearing rolling off a table. The table was 80cm high. The ball bearing landed 64cm from the edge of the table and I have to work out how long it took to do so.

vi=0
a=-9.8ms-2
d=64cm
t=?

0-2.9-64/-9.8 = okay... the answer is meant to be 0.40 s

[?]

 

[hitssquad]

Average velocity equals half of terminal velocity

Originally posted by physicsgirl_nz
Okay, the problem involving the formula was a ball bearing rolling off a table. The table was 80cm high. The ball bearing landed 64cm from the edge of the table and I have to work out how long it took to do so.

vi=0
a=-9.8ms-2
d=64cm
t=?

0-2.9-64/-9.8 = okay... the answer is meant to be 0.40 s

If you're only looking for the answer of how long it took for the bearing to travel from the very edge of the table to the floor, then it seems to me that the 64cm horizontal-distance figure would be extraneous to the problem (distractor information).


Since average vertical speed over the course of the fall is going to be half of the terminal speed (assuming constant downward acceleration), and the speed at any given moment is 9.8 m/s * the number of seconds travelled...
http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5b.html

...we can easily check the 0.40 s answer you came up with.


In .4 s, the bearing would have reached a terminal downward speed of .4 * 9.8 m/s = 3.92 m/s.
http://www.google.com/search?hl=en&ie=UTF-8&oe=UTF-8&q=.4+*+9.8+m/s

The average downward speed over that .4 s would be half of the terminal speed (since the acceleration was constant). .5 * 3.92 m/s = 1.96 m/s. Taking that average speed of 1.96 m/s and multiplying by .4 seconds gives us .4 * 1.96 = a distance fallen of 0.784 meters (78.4 centimeters).


Since the problem stated that the vertical distance fallen was .80 meters (80 centimeters), your answer of 0.40 s is pretty close.





-Chris

 

[physicsgirl_nz]

Thanks guys