Best 1/4mile with the 1.8 liter 10th gen.

[bus106]

Your best time manual,,, i made 16,7 sec @81.5mph

[buddy97]

i think there is only one way to get a good 1/4 mile time in a corolla:

[cabur]

Even going down straight Corolla will lose to others because it's lighter. Ha ha.

[johncal]

Originally Posted by cabur

Even going down straight Corolla will lose to others because it's lighter. Ha ha.

Funny.... but everything falls at the same speed regardless of weight. One of Newton's Laws. (Not the action and reaction one)

At least the Corolla wouldn't lose.

[cabur]

Will a feather fall at the same rate as a stone?
I think the formula is gravity X density.

[johncal]

Originally Posted by cabur

Will a feather fall at the same rate as a stone?
I think the formula is gravity X density.

Has nothing to do with density. In that case it's wind resistance. If you put a feater and a bowling ball in a vacuum tube, they will fall at the same rate. Here's something I pulled off the web. Couln't help myself............





If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. This statement follows from the law of conservation of energy and has been demonstrated experimentally by dropping a feather and a lead ball in an airless tube.

When air resistance plays a role, the shape of the object becomes important. In air, a feather and a ball do not fall at the same rate. In the case of a pen and a bowling ball air resistance is small compared to the force a gravity that pulls them to the ground. Therefore, if you drop a pen and a bowling ball you could probably not tell which of the two reached the ground first unless you dropped them from a very very high tower.

Answered by: Dr. Michael Ewart, Researcher at the University of Southern California


The above answer is perfectly correct, but, this is a question that confuses many people, and they are hardly satisfied by us self-assured physcists' answers. There is one good explanation which makes everybody content -- which does not belong to me, but to some famous scientist but I can't remember whom (Galileo?); and I think it would be good to have it up here.

(The argument has nothing to do with air resistance, it is assumed to be absent. The answer by Dr. Michael Ewart answers that part already.)

The argument goes as follows: Assume we have a 10kg ball and a 1kg ball. Let us assume the 10kg ball falls faster than the 1kg ball, since it is heavier. Now, lets tie the two balls together. What will happen then? Will the combined object fall slower, since the 1kg ball will hold back the 10kg ball? Or will the combination fall faster, since it is now an 11kg object? Since both can't happen, the only possibility is that they were falling at the same rate in the first place.

Sounds extremely convincing. But, I think there is a slight fallacy in the argument. It mentions nothing about the nature of the force involved, so it looks like it should work with any kind of force! However, it is not quite true. If we lived on a world where the 'falling' was due to electrical forces, and objects had masses and permanent charges, things would be different. Things with zero charge would not fall no matter what their mass is. In fact, the falling rate would be proportional to q/m, where q is the charge and m is the mass. When you tie two objects, 1 and 2, with charges q1, q2, and m1, m2, the combined object will fall at a rate (q1+q2)/(m1+m2). Assuming q1/m1 < q2/m2, or object 2 falls faster than object one, the combined object will fall at an intermediate rate (this can be shown easily). But, there is another point. The 'weight' of an object is the force acting on it. That is just proportional to q, the charge. Since what matters for the falling rate is q/m, the weight will have no definite relation to rate of fall. In fact, you could have a zero-mass object with charge q, which will fall infinitely fast, or an infinite-mass object with charge q, which will not fall at all, but they will 'weigh' the same! So, in fact, the original argument should be reduced to the following statement, which is more accurate:

If all objects which have equal weight fall at the same rate, then _all_ objects will fall at the same rate, regardless of their weight.

In mathematical terms, this is equivalent to saying that if q1=q2 then m1=m2 or, q/m is the same for all objects, they will all fall at the same rate! All in all, this is pretty hollow an argument.

Going back to the case of gravity.. The gravitational force is



( G is a constant, called constant of gravitation, M is the mass of the attracting body (here, earth), and m1 is the 'gravitational mass' of the object.)

And newton's law of motion is



where m2 is the 'inertial mass' of the object, and a is the acceleration.

Now, solving for acceleration, we find:



Which is proportional to m1/m2, i.e. the gravitational mass divided by the inertial mass. This is our old 'q/m' from the electrical case! Now, if and only if m1/m2 is a constant for all objects, (this constant can be absorbed into G, so the question can be reduced to m1=m2 for all objects) they will all fall at the same rate. If this ratio varies, then we will have no definite relation between rate of fall, and weight.

So, all in all, we are back to square one. Which is just canceling the masses in the equations, thus showing that they must fall at the same rate. The equality of the two masses is a necessity for general relativity, and enters it naturally. Also, the two masses have been found to be equal to extremely good precision experimentally. The correct answer to the question 'why objects with different masses fall at the same rate?' is, 'beacuse the gravitational and inertial masses are equal for all objects.'

Then, why does the argument sound so convincing? Since our daily experience and intuition dictates that things which weigh the same, fall at the same rate. Once we assume that, we have implicitly already assumed that the gravitational mass is equal to the inertial mass. (Wow, what things we do without noticing!). The rest of the argument follows easily and naturally...

Answered by: Yasar Safkan, Physics Ph.D. Candidate, M.I.T.




So at least I know the Corolla won't lose.... johncal

[trucknut1]

^^ Oh John.... ^^

now what i am wondering is if the corolla will fall faster in a 1/4 mile then in can accelerate??

[johncal]

Originally Posted by trucknut1

^^ Oh John.... ^^

now what i am wondering is if the corolla will fall faster in a 1/4 mile then in can accelerate??

I'm going to let you figure this one out. Here's the formula......

Let us know when you got it!




Kinematic Equations and Free Fall

As mentioned , a free-falling object is an object which is falling under the sole influence of gravity. That is to say that any object which is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s.
Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations which describe any object's motion are:

The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity.
There are a few conceptual characteristics of free fall motion which will be of value when using the equations to analyze free fall motion. These concepts are described as follows:

• An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object.
• If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.
• If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s.
• If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity which it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

These four principles and the four kinematic equations can be combined to solve problems involving the motion of free falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem-solving strategy which was introduced earlier in this lesson will be utilized.

Example A

Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below.
Diagram:

Given:

vi = 0.0 m/s d = -8.52 m
a = - 9.8 m/s2
Find:

t = ??
The next step involves identifying a kinematic equation which allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation which contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation which has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables.
Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.
-8.52 m = (0 m/s)*(t) + 0.5*(-9.8 m/s2)*(t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2)*(t)2
-8.52 m = (-4.9 m/s2)*(t)2
(-8.52 m)/(-4.9 m/s2) = t2
1.739 s2 = t2
t = 1.32 s
The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!


Example B

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below.
Diagram:

Given:

vi = 26.2 m/s vf = 0 m/s
a = -9.8 m/s2
Find:

d = ??
The next step involves identifying a kinematic equation which would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation which contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables.
Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.
(0 m/s)2 = (26.2 m/s)2 + 2*(-9.8m/s2)*d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2)*d
(-19.6 m/s2)*d = 0 m2/s2 -686.44 m2/s2
(-19.6 m/s2)*d = -686.44 m2/s2
d = (-686.44 m2/s2)/ (-19.6 m/s2)
d = 35.0 m
The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)
The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles. The next part of Lesson 6 provides a wealth of practice problems with answers and solutions.





In case you gave up, the answer is ........... 0 to 1320 ft takes 10 seconds @ 120 MPH (max speed). So it can free fall faster than it can accelerate.

I cheated. I used a sky diving calculator.

[jlauden]

On the other hand, if you figure our little cars have less drag than bigger cars (if they're pointed in the right direction anyway: down, in this case) we can assume that it would actually beat several less aerodynamic cars to the ground. Hooray Toyota!

[trucknut1]

Originally Posted by johncal

I'm going to let you figure this one out. Here's the formula......

Let us know when you got it!

In case you gave up, the answer is ........... 0 to 1320 ft takes 10 seconds @ 120 MPH (max speed). So it can free fall faster than it can accelerate.

I cheated. I used a sky diving calculator.

LMAO! I was actually trying to remember my high school physics classes as i am reading through it.. oh man my head hurts and i now also feel old. What a great thread

[sinnaever]

Not the best thread to read after a late friday night....

interesting... but WTF