Cain242
Welcome! edit
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A tag has been placed on User:Cain242, requesting that it be speedily deleted from Wikipedia. This has been done under section G11 of the criteria for speedy deletion, because the page seems to be unambiguous advertising which only promotes a company, product, group, service or person and would need to be fundamentally rewritten in order to become encyclopedic. Please read the guidelines on spam and Wikipedia:FAQ/Organizations for more information.
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Your submission at Articles for creation: sandbox (February 1) edit
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Hello! Cain242,
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February 2017 edit
Articles for creation/Redirects edit
Hi Cain242, I'm KGirlTrucker81, and I noticed you made an edit to Wikipedia:Articles for creation/Redirects. Your request was either empty, had elements missing, or the edit did not belong at WP:AFC/R, so for now it has been removed. If you would just like to experiment, please use the sandbox instead. When you feel ready to file a new request at Articles for creation/Redirects, you can simply click one of these buttons:
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Hi! edit
Hi Cain 242. It seems that you've been making some unconstructive edits to Wikipedia. A few funny unconstructive edits are fine, but please, for every unconstructive edit, make 2 constructive edits! Thanks, I really appreciate it! Ethanbas (talk) 16:05, 15 February 2017 (UTC)
i was thinking what i was adding was helpful The kinematic formulas are a set of formulas that relate the five kinematic variables listed below.
\Delta x\quad\text{Displacement} ΔxDisplacementdelta, x, space, D, i, s, p, l, a, c, e, m, e, n, t t\qquad\text{Time interval}~ tTime interval t, space, T, i, m, e, space, i, n, t, e, r, v, a, l, space v_0 ~~\quad\text{Initial velocity}~v 0 Initial velocity v, start subscript, 0, end subscript, space, space, space, I, n, i, t, i, a, l, space, v, e, l, o, c, i, t, y, space v\quad Cain242 (talk)\text{Final velocity}~v Final velocity v, space, space, space, space, F, i, n, a, l, space, v, e, l, o, c, i, t, y, space a \quad~~ \text{ Constant acceleration}~a Constant acceleration a, space, space, space, space, C, o, n, s, t, a, n, t, space, a, c, c, e, l, e, r, a, t, i, o, n, space [Why is the time interval now written as t?] tt\Delta xdelta, x\Delta tdelta, t\Deltadeltatt tt\Delta tdelta, t If we know three of these five kinematic variables—\Delta x, t, v_0, v, aΔx,t,v 0 ,v,adelta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a—for an object under constant acceleration, we can use a kinematic formula, see below, to solve for one of the unknown variables. The kinematic formulas are often written as the following four equations. [Where did these formulas come from?] \Large 1. \quad v=v_0+at1.v=v 0 +at1, point, space, v, equals, v, start subscript, 0, end subscript, plus, a, t \Large 2. \quad {\Delta x}=(\dfrac{v+v_0}{2})t2.Δx=( 2 v+v 0 )t2, point, space, delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t \Large 3. \quad \Delta x=v_0 t+\dfrac{1}{2}at^23.Δx=v 0 t+ 2 1 at 2 3, point, space, delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript \Large 4. \quad v^2=v_0^2+2a\Delta x4.v 2 =v 0 2 +2aΔx4, point, space, v, start superscript, 2, end superscript, equals, v, start subscript, 0, end subscript, start superscript, 2, end superscript, plus, 2, a, delta, x Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing. Also, the kinematic formulas assume all variables are referring to the same direction: horizontal xxx, vertical yyy, etc. [Wait, what?] v_{0x}
v, start subscript, 0, x, end subscript\Delta x, v_x, a_x delta, x, comma, v, start subscript, x, end subscript, comma, a, start subscript, x, end subscript
v_{0y}
v, start subscript, 0, y, end subscript\Delta y, v_y, a_y delta, y, comma, v, start subscript, y, end subscript, comma, a, start subscript, y, end subscript
xx v_x=v_{0x}+a_xt
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript, plus, a, start subscript, x, end subscript, t
\Delta x=v_{0x} t+\dfrac{1}{2}a_xt^2
delta, x, equals, v, start subscript, 0, x, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, x, end subscript, t, start superscript, 2, end superscript
v_x^2=v_{0x}^2+2a_x\Delta x
v, start subscript, x, end subscript, start superscript, 2, end superscript, equals, v, start subscript, 0, x, end subscript, start superscript, 2, end superscript, plus, 2, a, start subscript, x, end subscript, delta, x
\dfrac{v_x+v_{0x}}{2}=\dfrac{\Delta x}{t}
start fraction, v, start subscript, x, end subscript, plus, v, start subscript, 0, x, end subscript, divided by, 2, end fraction, equals, start fraction, delta, x, divided by, t, end fraction
What is a freely flying object—i.e., a projectile? It might seem like the fact that the kinematic formulas only work for time intervals of constant acceleration would severely limit the applicability of these formulas. However one of the most common forms of motion, free fall, just happens to be constant acceleration. All freely flying objects—also called projectiles—on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude g=9.81\dfrac{\text{m}}{\text{s}^2}g=9.81 s 2 m g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction. \Large g=9.81\dfrac{\text{m}}{\text{s}^2}\quad \text{(Magnitude of acceleration due to gravity)}g=9.81 s 2 m (Magnitude of acceleration due to gravity)g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, space, left parenthesis, M, a, g, n, i, t, u, d, e, space, o, f, space, a, c, c, e, l, e, r, a, t, i, o, n, space, d, u, e, space, t, o, space, g, r, a, v, i, t, y, right parenthesis A freely flying object is defined as any object that is accelerating only due to the influence of gravity. We typically assume the effect of air resistance is small enough to ignore, which means any object that is dropped, thrown, or otherwise flying freely through the air is typically assumed to be a freely flying projectile with a constant downward acceleration of magnitude g=9.81\dfrac{\text{m}}{\text{s}^2}g=9.81 s 2 m g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction. This is both strange and lucky if we think about it. It's strange since this means that a large boulder will accelerate downwards with the same acceleration as a small pebble, and if dropped from the same height, they would strike the ground at the same time. [How can this be so?] It's lucky since we don't need to know the mass of the projectile when solving kinematic formulas since the freely flying object will have the same magnitude of acceleration, g=9.81\dfrac{\text{m}}{\text{s}^2}g=9.81 s 2 m g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, no matter what mass it has—as long as air resistance is negligible. Note that g=9.81\dfrac{\text{m}}{\text{s}^2}g=9.81 s 2 m g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction is just the magnitude of the acceleration due to gravity. If upward is selected as positive, we must make the acceleration due to gravity negative a_y=-9.81\dfrac{\text{m}}{\text{s}^2}a y =−9.81 s 2 m a, start subscript, y, end subscript, equals, minus, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction for a projectile when we plug into the kinematic formulas. Warning: Forgetting to include a negative sign is one of the most common sources of error when using kinematic formulas. How do you select and use a kinematic formula? We choose the kinematic formula that includes both the unknown variable we're looking for and three of the kinematic variables we already know. This way, we can solve for the unknown we want to find, which will be the only unknown in the formula. For instance, say we knew a book on the ground was kicked forward with an initial velocity of v_0=5\text{ m/s}v 0 =5 m/sv, start subscript, 0, end subscript, equals, 5, space, m, slash, s, after which it took a time interval t=3\text{ s}t=3 st, equals, 3, space, s for the book to slide a displacement of \Delta x=8\text{ m}Δx=8 mdelta, x, equals, 8, space, m. We could use the kinematic formula \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript to algebraically solve for the unknown acceleration aaa of the book—assuming the acceleration was constant—since we know every other variable in the formula besides aaa—\Delta x, v_0, tΔx,v 0 ,tdelta, x, comma, v, start subscript, 0, end subscript, comma, t. Problem solving tip: Note that each kinematic formula is missing one of the five kinematic variables—\Delta x, t, v_0, v, aΔx,t,v 0 ,v,adelta, x, comma, t, comma, v, start subscript, 0, end subscript, comma, v, comma, a. 1.v=v0+at(This formula is missing Δx.) 2.Δx=(v+v02)t(This formula is missing a.) 3.Δx=v0t+12at2(This formula is missing v.) 4.v2=v20+2aΔx(This formula is missing t.) To choose the kinematic formula that's right for your problem, figure out which variable you are not given and not asked to find. For example, in the problem given above, the final velocity vvv of the book was neither given nor asked for, so we should choose a formula that does not include vvv at all. The kinematic formula \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript is missing vvv, so it's the right choice in this case to solve for the acceleration aaa. [Shouldn't there be a fifth kinematic formula that is missing the initial velocity?] 5.Δx=vt−12at2(This formula is missing v0.) \Delta x=v_0 t+\dfrac{1}{2}at^2
delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscriptv_0 v, start subscript, 0, end subscriptvv
How do you derive the first kinematic formula, v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t ? This kinematic formula is probably the easiest to derive since it is really just a rearranged version of the definition of acceleration. We can start with the definition of acceleration, a=\dfrac{\Delta v}{\Delta t}a= Δt Δv a, equals, start fraction, delta, v, divided by, delta, t, end fraction \quadspace [Isn't this the average acceleration?] \Delta tdelta, ta=\dfrac{\Delta v}{\Delta t}
a, equals, start fraction, delta, v, divided by, delta, t, end fraction
aaa_{avg}
a, start subscript, a, v, g, end subscripta_{inst}
a, start subscript, i, n, s, t, end subscripta_{avg}
a, start subscript, a, v, g, end subscriptaa
aa Now we can replace \Delta vΔvdelta, v with the definition of change in velocity v-v_0v−v 0 v, minus, v, start subscript, 0, end subscript. a=\dfrac{v_-v_0}{\Delta t}a= Δt v − v 0 a, equals, start fraction, v, start subscript, minus, end subscript, v, start subscript, 0, end subscript, divided by, delta, t, end fraction Finally if we just solve for vvv we get v=v_0+a\Delta tv=v 0 +aΔtv, equals, v, start subscript, 0, end subscript, plus, a, delta, t And if we agree to just use ttt for \Delta tΔtdelta, t, this becomes the first kinematic formula. \Large v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t How do you derive the second kinematic formula, {\Delta x}=(\dfrac{v+v_0}{2})tΔx=( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t? A cool way to visually derive this kinematic formula is by considering the velocity graph for an object with constant acceleration—in other words, a constant slope—and starts with initial velocity v_0v 0 v, start subscript, 0, end subscript as seen in the graph below.
v~(m/s)v (m/s)t~(s)t (s)ttv_0v 0 vv The area under any velocity graph gives the displacement \Delta xΔxdelta, x. So, the area under this velocity graph will be the displacement \Delta xΔxdelta, x of the object. \Delta x=\text{ total area}Δx= total areadelta, x, equals, space, t, o, t, a, l, space, a, r, e, a We can conveniently break this area into a blue rectangle and a red triangle as seen in the graph above. The height of the blue rectangle is v_0v 0 v, start subscript, 0, end subscript and the width is ttt, so the area of the blue rectangle is v_0tv 0 tv, start subscript, 0, end subscript, t. The base of the red triangle is ttt and the height is v-v_0v−v 0 v, minus, v, start subscript, 0, end subscript, so the area of the red triangle is \dfrac{1}{2}t(v-v_0) 2 1 t(v−v 0 )start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis. The total area will be the sum of the areas of the blue rectangle and the red triangle. \Delta x=v_0t+\dfrac{1}{2}t(v-v_0)Δx=v 0 t+ 2 1 t(v−v 0 )delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis If we distribute the factor of \dfrac{1}{2}t 2 1 tstart fraction, 1, divided by, 2, end fraction, t we get \Delta x=v_0t+\dfrac{1}{2}vt-\dfrac{1}{2}v_0tΔx=v 0 t+ 2 1 vt− 2 1 v 0 tdelta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, v, t, minus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t We can simplify by combining the v_0v 0 v, start subscript, 0, end subscript terms to get \Delta x=\dfrac{1}{2}vt+\dfrac{1}{2}v_0tΔx= 2 1 vt+ 2 1 v 0 tdelta, x, equals, start fraction, 1, divided by, 2, end fraction, v, t, plus, start fraction, 1, divided by, 2, end fraction, v, start subscript, 0, end subscript, t And finally we can rewrite the right hand side to get the second kinematic formula. \Large \Delta x=(\dfrac{v+v_0}{2})tΔx=( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t This formula is interesting since if you divide both sides by ttt, you get \dfrac{\Delta x}{t}=(\dfrac{v+v_0}{2}) t Δx =( 2 v+v 0 )start fraction, delta, x, divided by, t, end fraction, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis. This shows that the average velocity \dfrac{\Delta x}{t} t Δx start fraction, delta, x, divided by, t, end fraction equals the average of the final and initial velocities \dfrac{v+v_0}{2} 2 v+v 0 start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction. However, this is only true assuming the acceleration is constant since we derived this formula from a velocity graph with constant slope/acceleration. How do you derive the third kinematic formula, \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript? There are a couple ways to derive the equation \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript. There's a cool geometric derivation and a less exciting plugging-and-chugging derivation. We'll do the cool geometric derivation first. Consider an object that starts with a velocity v_0v 0 v, start subscript, 0, end subscript and maintains constant acceleration to a final velocity of vvv as seen in the graph below.
v~(m/s)v (m/s)t~(s)t (s)ttv_0v 0 vv Since the area under a velocity graph gives the displacement \Delta xΔxdelta, x, each term on the right hand side of the formula \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript represents an area in the graph above. The term v_0tv 0 tv, start subscript, 0, end subscript, t represents the area of the blue rectangle since A_{rectangle}=hwA rectangle =hwA, start subscript, r, e, c, t, a, n, g, l, e, end subscript, equals, h, w. The term \dfrac{1}{2}at^2 2 1 at 2 start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript represents the area of the red triangle since A_{triangle}=\dfrac{1}{2}bhA triangle = 2 1 bhA, start subscript, t, r, i, a, n, g, l, e, end subscript, equals, start fraction, 1, divided by, 2, end fraction, b, h. [Wait, how?] ttv-v_0
v, minus, v, start subscript, 0, end subscript
ata, tv-v_0=at
v, minus, v, start subscript, 0, end subscript, equals, a, t
A_{triangle}=\dfrac{1}{2}bh=\dfrac{1}{2}t(v-v_0)=\dfrac{1}{2}t(at)=\dfrac{1}{2}at^2
A, start subscript, t, r, i, a, n, g, l, e, end subscript, equals, start fraction, 1, divided by, 2, end fraction, b, h, equals, start fraction, 1, divided by, 2, end fraction, t, left parenthesis, v, minus, v, start subscript, 0, end subscript, right parenthesis, equals, start fraction, 1, divided by, 2, end fraction, t, left parenthesis, a, t, right parenthesis, equals, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript
That's it. The formula \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript has to be true since the displacement must be given by the total area under the curve. We did assume the velocity graph was a nice diagonal line so that we could use the triangle formula, so this kinematic formula—like all the rest of the kinematic formulas—is only true under the assumption that the acceleration is constant. Here's the alternative plugging-and-chugging derivation. The third kinematic formula can be derived by plugging in the first kinematic formula, v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t, into the second kinematic formula, \dfrac{\Delta x}{t}=\dfrac{v+v_0}{2} t Δx = 2 v+v 0 start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction. If we start with second kinematic formula \dfrac{\Delta x}{t}=\dfrac{v+v_0}{2} t Δx = 2 v+v 0 start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction and we use v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t to plug in for vvv, we get \dfrac{\Delta x}{t}=\dfrac{(v_0+at)+v_0}{2} t Δx = 2 (v 0 +at)+v 0 start fraction, delta, x, divided by, t, end fraction, equals, start fraction, left parenthesis, v, start subscript, 0, end subscript, plus, a, t, right parenthesis, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction We can expand the right hand side and get \dfrac{\Delta x}{t}=\dfrac{v_0}{2}+\dfrac{at}{2}+\dfrac{v_0}{2} t Δx = 2 v 0 + 2 at + 2 v 0 start fraction, delta, x, divided by, t, end fraction, equals, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction, plus, start fraction, a, t, divided by, 2, end fraction, plus, start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction Combining the \dfrac{v_0}{2} 2 v 0 start fraction, v, start subscript, 0, end subscript, divided by, 2, end fraction terms on the right hand side gives us \dfrac{\Delta x}{t}=v_0+\dfrac{at}{2} t Δx =v 0 + 2 at start fraction, delta, x, divided by, t, end fraction, equals, v, start subscript, 0, end subscript, plus, start fraction, a, t, divided by, 2, end fraction And finally multiplying both sides by the time ttt gives us the third kinematic formula. \Large \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript Again, we used other kinematic formulas, which have a requirement of constant acceleration, so this third kinematic formula is also only true under the assumption that the acceleration is constant. How do you derive the fourth kinematic formula, v^2=v_0^2+2a\Delta xv 2 =v 0 2 +2aΔxv, start superscript, 2, end superscript, equals, v, start subscript, 0, end subscript, start superscript, 2, end superscript, plus, 2, a, delta, x? To derive the fourth kinematic formula, we'll start with the second kinematic formula: {\Delta x}=(\dfrac{v+v_0}{2})tΔx=( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t We want to eliminate the time ttt from this formula. To do this, we'll solve the first kinematic formula, v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t, for time to get t=\dfrac{v-v_0}{a}t= a v−v 0 t, equals, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction. If we plug this expression for time ttt into the second kinematic formula we'll get {\Delta x}=(\dfrac{v+v_0}{2})(\dfrac{v-v_0}{a})Δx=( 2 v+v 0 )( a v−v 0 )delta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, left parenthesis, start fraction, v, minus, v, start subscript, 0, end subscript, divided by, a, end fraction, right parenthesis Multiplying the fractions on the right hand side gives {\Delta x}=(\dfrac{v^2-v_0^2}{2a})Δx=( 2a v 2 −v 0 2 )delta, x, equals, left parenthesis, start fraction, v, start superscript, 2, end superscript, minus, v, start subscript, 0, end subscript, start superscript, 2, end superscript, divided by, 2, a, end fraction, right parenthesis And now solving for v^2v 2 v, start superscript, 2, end superscript we get the fourth kinematic formula. \Large v^2=v_0^2+2a\Delta xv 2 =v 0 2 +2aΔxv, start superscript, 2, end superscript, equals, v, start subscript, 0, end subscript, start superscript, 2, end superscript, plus, 2, a, delta, x What's confusing about the kinematic formulas? People often forget that the kinematic formulas are only true assuming the acceleration is constant during the time interval considered. Sometimes a known variable will not be explicitly given in a problem, but rather implied with codewords. For instance, "starts from rest" means v_0=0v 0 =0v, start subscript, 0, end subscript, equals, 0, "dropped" often means v_0=0v 0 =0v, start subscript, 0, end subscript, equals, 0, and "comes to a stop" means v=0v=0v, equals, 0. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be g=9.81\dfrac{\text{m}}{\text{s}^2}g=9.81 s 2 m g, equals, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, so this acceleration will usually not be given explicitly in a problem but will just be implied for a freely flying object. People forget that all the kinematic variables—\Delta x, v_o, v, aΔx,v o ,v,adelta, x, comma, v, start subscript, o, end subscript, comma, v, comma, a—except for ttt can be negative. A missing negative sign is a very common source of error. If upward is assumed to be positive, then the acceleration due to gravity for a freely flying object must be negative: a_g=-9.81\dfrac{\text{m}}{\text{s}^2}a g =−9.81 s 2 m a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction. The third kinematic formula, \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript, might require the use of the quadratic formula, see solved example 3 below. People forget that even though you can choose any time interval during the constant acceleration, the kinematic variables you plug into a kinematic formula must be consistent with that time interval. In other words, the initial velocity v_0v 0 v, start subscript, 0, end subscript has to be the velocity of the object at the initial position and start of the time interval ttt. Similarly, the final velocity vvv must be the velocity at the final position and end of the time interval ttt being analyzed. What do solved examples involving the kinematic formulas look like? Example 1: First kinematic formula, v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t
A water balloon filled with Kool-Aid is dropped from the top of a very tall building. What is the velocity of the water balloon after falling for t=2.35 \text{ s}t=2.35 st, equals, 2, point, 35, space, s? Assuming upward is the positive direction, our known variables are v_0=0 \quadv 0 =0v, start subscript, 0, end subscript, equals, 0, space (Since the water balloon was dropped, it started at rest.) t=2.35\text{ s} \quadt=2.35 st, equals, 2, point, 35, space, s, space (This is the time interval after which we want to find the velocity.) a_g=-9.81\dfrac{\text{m}}{\text{s}^2} \quada g =−9.81 s 2 m a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, space(This is implied since the water balloon is a freely falling object.) [Isn't the final velocity zero since it hits the ground?] 2.35 \text{ s}2, point, 35, space, s a_g=-9.81\dfrac{\text{m}}{\text{s}^2}
a, start subscript, g, end subscript, equals, minus, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction
The motion is vertical in this situation, so we'll use yyy as our position variable instead of xxx. The symbol we choose doesn't really matter as long as we're consistent, but people typically use yyy to indicate vertical motion. Since we don't know the displacement \Delta yΔydelta, y and we weren't asked for the displacement \Delta yΔydelta, y, we'll use the first kinematic formula v=v_{0}+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t, which is missing \Delta yΔydelta, y. v=v0+at(Use the first kinematic formula since it's missing Δy.) v=0\text{ m/s} +(-9.81\dfrac{\text{m}}{\text{s}^2})(2.35\text{ s}) \quad \text{(Plug in known values.)}v=0 m/s+(−9.81 s 2 m )(2.35 s)(Plug in known values.)v, equals, 0, space, m, slash, s, plus, left parenthesis, minus, 9, point, 81, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, right parenthesis, left parenthesis, 2, point, 35, space, s, right parenthesis, space, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis v=-23.1 \text { m/s}\quad \text{(Calculate and celebrate!)}v=−23.1 m/s(Calculate and celebrate!) Note: The final velocity was negative since the water balloon was heading downward. [Can't we call downward the positive direction?] +9.8\dfrac{\text{m}}{\text{s}^2}
plus, 9, point, 8, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction
Example 2: Second kinematic formula, {\Delta x}=(\dfrac{v+v_0}{2})tΔx=( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t
A leopard is running at 6.20 m/s and after seeing a mirage that's taken the form of an ice cream truck; the leopard then speeds up to 23.1 m/s in a time of 3.3 s. How much ground did the leopard cover in going from 6.20 m/s to 23.1 m/s? Assuming the initial direction of travel is the positive direction, our known variables are v_0= 6.20\text{ m/s} \quadv 0 =6.20 m/sv, start subscript, 0, end subscript, equals, 6, point, 20, space, m, slash, s, space (The initial speed of the leopard) v= 23.1\text{ m/s} \quadv=23.1 m/sv, equals, 23, point, 1, space, m, slash, s, space (The final speed of the leopard) t=3.30\text{ s} \quadt=3.30 st, equals, 3, point, 30, space, s, space (The time it took for the leopard to speed up) Since we do not know the acceleration aaa and were not asked for the acceleration, we'll use the second kinematic formula for the horizontal direction {\Delta x}=(\dfrac{v+v_{0}}{2})tΔx=( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t, which is missing aaa. Δx=(v+v02)t(Use the second kinematic formula since it's missing a.) {\Delta x}=(\dfrac{23.1\text{ m/s}+6.20\text{ m/s}}{2})(3.30\text{ s} ) \quad \text{(Plug in known values.)}Δx=( 2 23.1 m/s+6.20 m/s )(3.30 s)(Plug in known values.)delta, x, equals, left parenthesis, start fraction, 23, point, 1, space, m, slash, s, plus, 6, point, 20, space, m, slash, s, divided by, 2, end fraction, right parenthesis, left parenthesis, 3, point, 30, space, s, right parenthesis, space, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis \Delta x=48.3 \text{ m} \quad \text{(Calculate and celebrate!)}Δx=48.3 m(Calculate and celebrate!) Example 3: Third kinematic formula, \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, start superscript, 2, end superscript
A student is fed up with doing her kinematic formula homework, so she throws her pencil straight upward at 18.3 m/s. How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown? Assuming upward is the positive direction, our known variables are v_0=18.3 \text { m/s} \quadv 0 =18.3 m/sv, start subscript, 0, end subscript, equals, 18, point, 3, space, m, slash, s, space (The initial upward velocity of the pencil) \Delta y=12.2\text{ m} \quadΔy=12.2 mdelta, y, equals, 12, point, 2, space, m, space (We want to know the time when the pencil moves through this displacement.) a=-9.81\dfrac{\text{ m}}{\text{ s}^2} \quada=−9.81 s 2 m a, equals, minus, 9, point, 81, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, space (The pencil is a freely flying projectile.) Since we don't know the final velocity vvv and we weren't asked to find the final velocity, we will use the third kinematic formula for the vertical direction \Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2Δy=v 0y t+ 2 1 a y t 2 delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, start superscript, 2, end superscript, which is missing vvv. \Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2 \quad \text{(Start with the third kinematic formula.)}Δy=v 0y t+ 2 1 a y t 2 (Start with the third kinematic formula.)delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, start superscript, 2, end superscript, space, left parenthesis, S, t, a, r, t, space, w, i, t, h, space, t, h, e, space, t, h, i, r, d, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, point, right parenthesis Normally we would just solve our expression algebraically for the variable we want to find, but this kinematic formula can not be solved algebraically for time if none of the terms are zero. That's because when none of the terms are zero and ttt is the unknown variable, this equation becomes a quadratic equation. We can see this by plugging in known values. 12.2\text{ m}=(18.3\text{ m/s})t+\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2 \quad \text{(Plug in known values.)}12.2 m=(18.3 m/s)t+ 2 1 (−9.81 s 2 m )t 2 (Plug in known values.)12, point, 2, space, m, equals, left parenthesis, 18, point, 3, space, m, slash, s, right parenthesis, t, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, t, start superscript, 2, end superscript, space, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis To put this into a more solvable form of the quadratic equation, we move everything onto one side of the equation. Subtracting 12 m from both sides we get 0=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2+(18.3\text{ m/s})t -12.2\text{ m} \quad \text{(Put it into the form of the quadratic equation.)}0= 2 1 (−9.81 s 2 m )t 2 +(18.3 m/s)t−12.2 m(Put it into the form of the quadratic equation.)0, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, t, start superscript, 2, end superscript, plus, left parenthesis, 18, point, 3, space, m, slash, s, right parenthesis, t, minus, 12, point, 2, space, m, space, left parenthesis, P, u, t, space, i, t, space, i, n, t, o, space, t, h, e, space, f, o, r, m, space, o, f, space, t, h, e, space, q, u, a, d, r, a, t, i, c, space, e, q, u, a, t, i, o, n, point, right parenthesis At this point, we solve the quadratic equation for time ttt. The solutions of a quadratic equation in the form of at^2+bt+c=0at 2 +bt+c=0a, t, start superscript, 2, end superscript, plus, b, t, plus, c, equals, 0 are found by using the quadratic formula t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}t= 2a −b±√ b 2 −4ac t, equals, start fraction, minus, b, plus minus, square root of, b, start superscript, 2, end superscript, minus, 4, a, c, end square root, divided by, 2, a, end fraction. For our kinematic equation a=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})a= 2 1 (−9.81 s 2 m )a, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, b=18.3\text{ m/s}b=18.3 m/sb, equals, 18, point, 3, space, m, slash, s, and c=-12.2\text{ m}c=−12.2 mc, equals, minus, 12, point, 2, space, m. So, plugging into the quadratic formula, we get t=\dfrac{-18.3\text{ m/s}\pm\sqrt{(18.3\text{ m/s})^2-4[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})(-12.2\text{ m})]}}{2[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})]}t= 2[ 2 1 (−9.81 s 2 m )] −18.3 m/s±√ (18.3 m/s) 2 −4[ 2 1 (−9.81 s 2 m )(−12.2 m)] t, equals, start fraction, minus, 18, point, 3, space, m, slash, s, plus minus, square root of, left parenthesis, 18, point, 3, space, m, slash, s, right parenthesis, start superscript, 2, end superscript, minus, 4, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, left parenthesis, minus, 12, point, 2, space, m, right parenthesis, close bracket, end square root, divided by, 2, open bracket, start fraction, 1, divided by, 2, end fraction, left parenthesis, minus, 9, point, 81, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, close bracket, end fraction Since there is a plus or minus sign in the quadratic formula, we get two answers for the time ttt: one when using the ++plus and one when using the -−minus. Solving the quadratic formula above gives these two times: t=0.869\text{ s}t=0.869 st, equals, 0, point, 869, space, s and t=2.86\text{ s}t=2.86 st, equals, 2, point, 86, space, s There are two positive solutions since there are two times when the pencil was 12.2 m high. The smaller time refers to the time required to go upward and first reach the displacement of 12.2 m high. The larger time refers to the time required to move upward, pass through 12.2 m high, reach a maximum height, and then fall back down to a point 12.2 m high. So, to find the answer to our question of "How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?" we would choose the smaller time t=0.869\text{ s}t=0.869 st, equals, 0, point, 869, space, s. [What if the quadratic formula gives a negative answer?] Example 4: Fourth kinematic formula, v^2=v_0^2+2a\Delta xv 2 =v 0 2 +2aΔxv, start superscript, 2, end superscript, equals, v, start subscript, 0, end subscript, start superscript, 2, end superscript, plus, 2, a, delta, x
A European motorcyclist starts with a speed of 23.4 m/s and, seeing traffic up ahead, decides to slow down over a length of 50.2 m with a constant deceleration of magnitude 3.20\dfrac{\text{ m}}{\text{ s}^2}3.20 s 2 m 3, point, 20, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction. Assume the motorcycle is moving forward for the entire trip. What is the new velocity of the motorcyclist after slowing down through the 50.2 m? Assuming the initial direction of travel is the positive direction, our known variables are v_0=23.4 \text { m/s} \quadv 0 =23.4 m/sv, start subscript, 0, end subscript, equals, 23, point, 4, space, m, slash, s, space (The initial forward velocity of the motorcycle) a=-3.20\dfrac{\text{ m}}{\text{ s}^2} \quada=−3.20 s 2 m a, equals, minus, 3, point, 20, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, space (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.) \Delta x=50.2\text{ m} \quadΔx=50.2 mdelta, x, equals, 50, point, 2, space, m, space (We want to know the velocity after the motorcycle moves through this displacement.) Since we don't know the time ttt and we weren't asked to find the time, we will use the fourth kinematic formula for the horizontal direction v_x^2=v_{0x}^2+2a_x\Delta xv x 2 =v 0x 2 +2a x Δxv, start subscript, x, end subscript, start superscript, 2, end superscript, equals, v, start subscript, 0, x, end subscript, start superscript, 2, end superscript, plus, 2, a, start subscript, x, end subscript, delta, x, which is missing ttt. v_x^2=v_{0x}^2+2a_x\Delta x \quad \text{(Start with the fourth kinematic formula.)}v x 2 =v 0x 2 +2a x Δx(Start with the fourth kinematic formula.)v, start subscript, x, end subscript, start superscript, 2, end superscript, equals, v, start subscript, 0, x, end subscript, start superscript, 2, end superscript, plus, 2, a, start subscript, x, end subscript, delta, x, space, left parenthesis, S, t, a, r, t, space, w, i, t, h, space, t, h, e, space, f, o, u, r, t, h, space, k, i, n, e, m, a, t, i, c, space, f, o, r, m, u, l, a, point, right parenthesis v_x=\pm\sqrt{v_{0x}^2+2a_x\Delta x} \quad \text{(Algebraically solve for the final velocity.)}v x =±√ v 0x 2 +2a x Δx (Algebraically solve for the final velocity.)v, start subscript, x, end subscript, equals, plus minus, square root of, v, start subscript, 0, x, end subscript, start superscript, 2, end superscript, plus, 2, a, start subscript, x, end subscript, delta, x, end square root, space, left parenthesis, A, l, g, e, b, r, a, i, c, a, l, l, y, space, s, o, l, v, e, space, f, o, r, space, t, h, e, space, f, i, n, a, l, space, v, e, l, o, c, i, t, y, point, right parenthesis Note that in taking a square root, you get two possible answers: positive or negative. Since our motorcyclist will still be going in the direction of motion it started with and we assumed that direction was positive, we'll choose the positive answer v_x=+\sqrt{v_{0x}^2+2a_x\Delta x}v x =+√ v 0x 2 +2a x Δx v, start subscript, x, end subscript, equals, plus, square root of, v, start subscript, 0, x, end subscript, start superscript, 2, end superscript, plus, 2, a, start subscript, x, end subscript, delta, x, end square root. Now we can plug in values to get v_x=\sqrt{(23.4\text{ m/s})^2+2(-3.20\dfrac{\text{ m}}{\text{ s}^2})(50.2\text{ m})} \quad \text{(Plug in known values.)}v x =√ (23.4 m/s) 2 +2(−3.20 s 2 m )(50.2 m) (Plug in known values.)v, start subscript, x, end subscript, equals, square root of, left parenthesis, 23, point, 4, space, m, slash, s, right parenthesis, start superscript, 2, end superscript, plus, 2, left parenthesis, minus, 3, point, 20, start fraction, space, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, left parenthesis, 50, point, 2, space, m, right parenthesis, end square root, space, left parenthesis, P, l, u, g, space, i, n, space, k, n, o, w, n, space, v, a, l, u, e, s, point, right parenthesis v_x=15.0\text{ m/s} \quad \text{(Calculate and celebrate!)}v x =15.0 m/s(Calculate and celebrate!)
Talk:Joker edit
Cain242, I undid your edit at Talk:Joker. Edits to that page should be about improving or changing the page at Joker. I didn't understand why you wrote about quantum mechanics there. To get a better understanding of how Wikipedia works, you might want to read and ask questions at the Teahouse or Wikipedia:Contributing to Wikipedia. Thank you. SchreiberBike | ⌨ 14:27, 16 February 2017 (UTC)
- I think its funny to have random things in random places plus it will teach people. :)
- I'll keep an eye on what you do and will remove things that don't make Wikipedia better. You can make Wikipedia (and the world) better by fixing errors you see or adding good material to articles. Unfortunately Wikipedia isn't here to make you laugh. I wish you the best. SchreiberBike | ⌨ 22:09, 16 February 2017 (UTC)
what does it matter where the info is as long as it is teaching people it should not matter wikipedia is here for knowledge thats what i put here is positive info