Help me with my test!?

Question

Hi, I'm in Year 9 at High School (New Zealand) and I got a Science test coming up. I had parent teacher inteviews and I learned I'm failing Science. Now tommorrow I've got a test on Physics. I need to know how to use a force meter, (Newtons) and how to do equation involving things like Speed, distance and force ect.

Answer 1

A newton is the amount of force required to accelerate a mass of one kilogram at a rate of one meter per sec per sec.
Description of Motion in One Dimension
Motion is described in terms of displacement (x), time (t), velocity (v), and acceleration (a). Velocity is the rate of change of displacement and the acceleration is the rate of change of velocity.
In order to derive new equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity, displacement, and time. There are three ways to pair them up: velocity-time, displacement-time, and velocity-displacement. In this order, they are also often called the first, second, and third equations of motion, but there is no compelling reason to learn these names. Since we are dealing with motion in a straight line, the symbol x will be used for displacement. Direction will be indicated by the sign (positive quantities point in +x direction, while negative quantities point in the -x direction). Determining which direction is positive and which negative is entirely arbitrary. The laws of physics are isotropic; that is, they are independent of the orientation of the coordinate system. As long as you are consistent, it doesn't matter. Some problems are easier to understand and solve, however, when one direction is chosen positive over another.

velocity-time
The relation between velocity and time is a simple one during constantly accelerated, straight-line motion. Constant acceleration implies a uniform rate of change in the velocity. The longer the acceleration, the greater the change in velocity. If after a time velocity increases by a certain amount, after twice that time it should increase by twice that amount. Change in velocity is directly proportional to time when acceleration is constant. If an object already started with a certain velocity, then its new velocity would be the old velocity plus this change. You ought to be able to see the equation in your mind's eye already.

This is the easiest of the three equations to derive formally. Start from the definition of acceleration, expand the Δv term, and solve for v as a function of t.


a = Δv = v − v0

Δt Δt

v = v0 + aΔt (1)


Since acceleration is also the first derivative of velocity with respect to time, this equation can also be derived using calculus. Just reverse the action of the definition. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. Since acceleration is assumed constant, this is quite easy.


a = dv

dt
dv = a dt
v Δt
∫ dv = ∫ a dt
v0 0
v − v0 = aΔt
v = v0 + aΔt (1)


The symbol v0 (v nought) is called the initial velocity. It is often thought of as the "first velocity" but this is a rather naïve way to describe it. Take the case of a meteor hurtling towards the earth. What is its initial velocity? If you want v0 to be the first velocity, then the problem is over before it started. Who could possibly say what a meteor's velocity was at its birth? There is no way to answer this question. A better definition would be to say that an initial velocity is the velocity that a moving object has when it first becomes important in a problem. Say the meteor was spotted deep in space and the problem was to determine its trajectory, then the initial velocity would be the velocity at the time it was observed. But if the problem were to determine its velocity on impact, then it's initial velocity would more likely be the velocity it had when it entered the earth's atmosphere. In this case, the answer to, "What's the initial velocity?" is "It depends". This turns out to be the answer to a lot of questions.

The symbol v is then the velocity some time Δt after the initial velocity. It is often called the final velocity but this does not make it an object's "last velocity". Take the case of the meteor. What velocity is represented by the symbol v? If you've been paying attention, then you should have anticipated the answer. It depends. It could be the velocity of the meteor as it passes by the moon, as it enters the earth's atmosphere, or as it strikes the earth's surface. It could also be the meteorite's velocity as it sits in the bottom of a crater (zero). Is this then the final velocity? Who knows. Someone could extract the meteorite from its hole in the ground and drive away with it. Is this relevant? Well, maybe. It depends. There's no rule for this kind of thing. You'll have to understand the problem and then make a decision for yourself.

The last part of this equation aΔt is the change in the velocity from the initial value. Recall that a is the rate of change of velocity and that Δt is the time interval since the object had its initial velocity v0. Rate multiplied by time equals change. Thus if an object were accelerating at 10 m/s2, after 5 s it would be moving 50 m/s faster than it was initially. If it started with a velocity of 15 m/s, its velocity after 5 s of acceleration would be 65 m/s.

displacement-time
The displacement of a moving object is directly proportional to both velocity and time. Move faster. Go farther. Move longer (as in longer time). Go farther. Acceleration compounds this simple situation. Now the velocity is also directly proportional to time. Try saying this in words and it sounds ridiculous. "Displacement is directly proportional to time and directly proportional to velocity, which is directly proportional to time." Time is a factor twice, making displacement proportional to the square of time. A car accelerating for two seconds would cover four times the distance of a car accelerating for only one second (22 = 4). A car accelerating for three seconds would cover nine times the distance (32 = 9).

Would that it should be so simple. This example only works when initial velocity is zero.vChange in displacement is proportional to the square of time when acceleration is constant and initial velocity is zero. A true general statement would have to take into account any initial velocity and how the velocity was changing. This results in a terribly messy proportionality statement. Change in displacement is directly proportional to time and proportional to the square of time when acceleration is constant. A function that is both linear and square is said to be quadratic, which allows us to compact the previous statement considerably. Change in displacement is a quadratic function of time when acceleration is constant

Proportionality statements are useful, but not as concise as equations. We still don't know what the constants of proportionality are for this problem. The only way to answer that is through algebra.

Start with the definition of velocity, expand Δx, and solve it for displacement.


v = Δx

Δt
Δx = vΔt
x − x0 = vΔt
x = x0 + vΔt (a)


To continue, we need to resort to a little trick first published in the Fourteenth Century at Merton College, Oxford (and sometimes called the Merton Rule). When acceleration is constant, the velocity will change uniformly from its initial value to its final value and the average will lie halfway between the extremes. Thus, the average velocity is just the arithmetic mean of the initial and final velocities. Average velocity is the average of the final and initial velocities when acceleration is constant.


v = v + v0 (4)

2


Substitute the first equation of motion (1) into this equation (4) and simplify with the intent of eliminating v.


v = v + v0 = (v0 + aΔt ) + v0 = 2v0 + aΔt = v0 + ½ aΔt (b)

2 2 2


Finally, substitute (b) into (a) and solve for x as a function of t.


x = x0 + vΔt
x = x0 + (v0 + aΔt)Δt
x = x0 + v0Δt + ½ aΔt2 (2)


Since velocity is also the first derivative of displacement with respect to time, this equation can also be derived using calculus. In fact, it's much easier than using algebra. Just reverse the action of the definition. Instead of differentiating displacement to find velocity, integrate velocity (1) to find displacement.


v = dx

dt
dx = v dt
x Δt Δt
∫ dx = ∫ v dt = ∫ (v0 + aΔt )dt
x0 0 0
x − x0 = v0Δt + ½ aΔt2
x = x0 + v0Δt + ½ aΔt2 (2)


The symbol x0 (ex nought) is the initial displacement. Many times, this value is zero and if it isn't, we can make it so. If you ask me, "When should we do this?" I would say, "It depends on the problem," and leave it to you to decide. There is no rule that you can memorize is this case. You have to understand what the equation says and then learn how to apply it to a particular situation. Similarly x is often called the final displacement, but this does not make it the "last displacement", rather it is the displacement at the end of the time interval during which the acceleration was constant.

Something else to notice is the similarity between equations (2) and (a). When acceleration is zero, our second equation of motion reverts to a rearranged constant velocity equation. As predicted, displacement is in part directly proportional to time and in part directly proportional to time squared.

x = x0 + v0Δt + ½ aΔt2 (2)
x = x0 + vΔt (a)

Although the velocity symbols in the two equations may look different, they do indeed represent the same quantity. If there is no acceleration, then the velocity is constant, which means that the initial velocity is the same as the final velocity is the same as the average velocity. The acceleration term at the end is an adjustment to the constant velocity equation to account for the the fact that the velocity is changing. A positive acceleration would increase the displacement and a negative acceleration would decrease it. This is exactly what one would expect. If an object's velocity was increasing, it would move farther than if it had stayed at a constant velocity. Likewise, if an object's velocity was decreasing, it would have a smaller displacement than if its velocity were constant. It's good to see that the equations behave in a realistic manner. Otherwise all this math would be a waste of time.

velocity-displacement
We have just seen that velocity is directly proportional to time and displacement is proportional to time squared. With a little bit of thinking, a new proportionality statement should be apparent. Change in displacement is proportional to the change in the square of velocity when acceleration is constant. This statement is particularly important for driving safety. When you double the speed of a car, it takes four times the distance to stop it. Triple the speed and you'll need nine times the distance. Like the previous relationships, it also depends on initial velocity. Unfortunately determining the effect using reasoning alone is a real chore. Do the algebra instead.

The last two equations each described one kinematic variable as a function of time. It would be nice if we also had an equation that was independent of time. That is, we want to answer the question, "What is the relationship between velocity and displacement?" The method of doing this should be readily apparent. We've got to combine our first two equations of motion together in a manner that will eliminate time as a variable. The easiest way to do that is to solve one equation for time and then substitute it into the other. The second equation of motion is a quadratic and solving it for time would introduce a lot of nastiness into the algebra. It should be apparent that solving the first equation of motion (1) for time and substituting it into the second (2) will be the easier process.


x = x0 + v0Δt + ½ aΔt2 (1) v = v0 + aΔt (2)
x − x0 = v0Δt + ½ aΔt2 ← Δt = v − v0

a
x − x0 = v0 ( v − v0 ) + 1 a ( v − v0 )2

a 2 a
x − x0 = vv0 − v02 + v2 − 2vv0 + v02

a 2a
2a(x − x0) = (2vv0 − 2v02) + (v2 − 2vv0 + v02)
2a(x − x0) = v2 − v02
v2 = v02 + 2a(x − x0) (3)


As expected, displacement is proportional to velocity squared. (Initial velocity squared is just a constant that must be dealt with.) Unlike the first and second equations of motion, there is no advantage in using calculus to derive the third equation of motion. Algebra is the way to go.

Summary
The equations of motion are valid only when …
acceleration is constant and
motion is constrained to a straight line.
Some proportionality statements:
Change in velocity is directly proportional to time.
Change in displacement is …
proportional to the square of time when initial velocity is zero, or
directly proportional to time and proportional to the square of time, or
a quadratic function of time.
Change in displacement is proportional to the change in the square of velocity.
Average velocity is the average of the final and initial velocities when acceleration is constant.
The one dimensional equations of motion for constant acceleration are …
Equations of Motion
relationship equation
(1) velocity-time v = v0 + at
(2) displacement-time x = x0 + v0t + ½at2
(3) velocity-displacemen v2 = v02 + 2a(x − x0)
(4) average velocity v = ½(v + v0)


Problems
practice
Cars cruise down an expressway at 25 m/s. Engineers want to design an interchange for a deceleration of -2.0 m/s2 over 3.0. s.
What velocity will cars have at the end of the approach?
What minimum approach length will satisfy these requirements?
What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)
Solutions …
What velocity will cars have at the end of the approach?
v0 = 25 m/s v = v0 + at
a = -2.0 m/s2 v = (25 m/s) + (-2.0 m/s2)(3.0 s)
Δt = 3.0 s v = 19 m/s
v = ??


What minimum approach length will satisfy these requirements?
v0 = 25 m/s Δx = v0t + ½at2
a = -2.0 m/s2 Δx = (25 m/s)(3.0 s) + ½(-2.0 m/s2)(3.0 s)2
Δt = 3.0 s Δx = 66 m
Δx = ??


What maximum velocity could a car entering the interchange have and still be able to exit at the intended velocity? (Assume an extreme deceleration of four times the usual rate.)
v = 19 m/s v2 = v02 + 2aΔx
a = -8.0 m/s2 v02 = v2 − 2aΔx
Δx = 66 m v02 = (19 m/s)2 + 2(-8.0 m/s2)(66 m)
v0 = ?? v0 = 38 m/s


A car with an initial velocity of 60 mph needs 144 feet to come to a complete stop. Determine the stopping distance of this same car, under identical conditions with an initial velocity of 30 mph, 20 mph, and 10 mph.
The hard way to solve this problem is to do it the way that many students think is the easy way. That is, the "plug and chug" method shown in the previous examples. This method appears easy since it requires very little thought, but it turns out to be quite demanding.

Answer 2

Equation 1
S=ut +at^2/2

Equation 2
F=ma

where
u=initial speed
a=acceleration
t=time taken
S=distance
F=Force
m=mass

Answer 3

I hope you will not go to your school tomorrow.

Answer 4

What he said.......