Hi guys this is just from p1 of my trig book, which I'm trying to relearn, but I'm already stumped. Please help me to remember how this makes sense: d= 16t^2 where d is in feet and t is in seconds. Because as I remember it, d= r*t So my question is in what circumstance can r ever equal t in order to make up t^2? Does this have to do with g (is it g?) being 10m/s^2 ? .......................................... The full text: When an object is dropped, the distance it travels depends upon the amount of time elpased since its release. For a given time t, the approx. dist d is given by the equation d = 16t^2, where d is in feet and t is in seconds. The solutions to the equation d=16t^2 can be written as ordered pairs (t,d), such as { (1,16), (2,64), (3,144) ...}. A set of ordered pairs of numbers is called a relation. The set of replacements for the first variable is called the domain of the relation.... ........................................ Sorry this is so basic but im really blank.
But a=g=9.8~10=/=16, so if his question is really a free falling problem, how come you can get d=16t^2 but not d=10t^2?
The text of her problem gives \(d=16t^2\), so it is not free falling in the Earth gravitational field but in a stronger field. BTW, if it were Earth, then the equation is \(d=\frac{gt^2}{2}\), so, it cannot be d=10t^2.
\(d = 16 \, \textrm{feet} \, \left( \frac{ t }{1 \, \textrm{second}} \right)^2 \\ d = k t^2 , \quad \quad \quad k = 16 \, \textrm{feet} \cdot \textrm{seconds}^{\tiny -2}\) This describes a particular relationship between d and t. By including the units and not just the numbers, one can ensure the relationship has the same physical interpretation by anyone and time and distance aren't interpreted as hours and kilometers, respectively. That is the equation for a constant rate of change of distance (speed). It only applies in the special case when the speed is constant. For falling objects, the speed gets faster as time passes. For example if an object falls distance D in the first second, it falls 3D in the next second, and 5D in the second after that for totals of D, 4D and 9D at the end of 1,2,and 3 seconds respectively. Here, on the surface of the Earth, D happens to be very close to 16 feet. The relationship for constant acceleration, a, and some starting velocity, u, is \(d = u t + \frac{1}{2} a t^2\) So \(a = g = 10 \, \textrm{m} \cdot \textrm{s}^{\tiny -2} = 32.8084 \, \textrm{feet} \cdot \textrm{s}^{\tiny -2}\) is just about the same thing (to rounding) as saying \(\frac{1}{2} a = \frac{1}{2} g = 16 \, \textrm{feet} \cdot \textrm{s}^{\tiny -2} = 4.8768 \, \textrm{m} \cdot \textrm{s}^{\tiny -2} \) . Of course in this example \(u = 0\). Thanks for this. Too many people cut out important parts of the context.
V=Final velocity, U=Initial velocity. Do you have a problem understanding what initial and final velocity are? The equations I posted are 100% accurate, so if you are saying something is nonsense it isn't the equations I posted, it's some other nonsense of yours. That is unless you care to backup your claim that something I posted is nonsense?
I didn't cut and paste anything. Some of those equations are well known, and some of those I worked out myself. Edit, Maybe you fail to see why there is 4 of each?
I didn't fail to notice anything, I gave the equations that are accurate. Your ignorance of the difference between initial and final velocity is laughable, and is causing you to make unsound judgements about me. Get a clue!
You copied a long list of generic equations. You failed to notice that the exercise requires zero initial speed. You claim that you "worked out" the equations by yourself. Your inability to do any math is well documented. OK, let's test that: how did you get \(d=vt-\frac{1}{2}at^2\). Let's see the steps, it is very simple math. Let's see if you can do it.
No free lessons for you. If you claim I copied and pasted, then post the link where I got them, otherwise you're just talking crap, as usual!
...yet no link. The only thing the reader is to assume is that you are making unsubstantiated assertions, ie talking crap! You like that "no free lessons" bit? I was actually laughing out loud!
You posted a JPEG, you are unable to do tex and you are unable to do math. Your inabilities are well documented.
Still no link. Maybe if AN would make a presence he could verify, because I specifically asked him to check my work, about a year ago. I made the tex here and made a pic of the tex. So you're back to where you started, talking crap! It always comes back to you talking crap!
So, you should have no difficulty in "working out" \(d=vt-\frac{1}{2}at^2\). But you can't. Your inability is well documented.
