describe the three derivation of motion
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constant acceleration
For the sake of accuracy, this section should be entitled "One dimensional equations of motion for constant acceleration". Given that such a title would be a stylistic nightmare, let me begin this section with the following qualification. These equations of motion are valid only when acceleration is constant and motion is constrained to a straight line.
Given that we live in a three dimensional universe in which the only constant is change, you may be tempted to dismiss this section outright. It would be correct to say that no object has ever traveled in a straight line and with a constant acceleration anywhere in the universe at any time — not today, not yesterday, not tomorrow, not five billion years ago, not thirty billion years in the future, never. This I can say with absolute
Our goal in this section, is to derive new equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity (v), position (s), and time (t). There are three ways to pair them up: velocity-time, position-time, and velocity-position. 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.
As long as you are consistent within a problem, it doesn't matter.
velocity-time
The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. 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.
a =
Δv
Δt
Expand Δv to v − v0 and condense Δt to t.
a =
v − v0
t
Then solve for v as a function of t.
v = v0 + at [1]
This is the first equation of motion. It's written like a polynomial — a constant term (v0) followed by a first order term (at). Since the highest order is 1, it's more correct to call it a linear function.
The symbol v0 [vee nought] is called the initial velocity or the velocity a time t = 0. It is often thought of as the "first velocity" but this is a rather naive way to describe it. 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.
The symbol v is the velocity some time t after the initial velocity. It is often called the final velocity but this does not make it an object's "last velocity".
The last part of this equation at is the change in the velocity from the initial value. Recall that ais the rate of change of velocity and that t is the time after some initial event. Rate times time is change. Given an object accelerating at 10 m/s2, after 5 s it would be moving 50 m/s faster. If it started with a velocity of 15 m/s, then its velocity after 5 s would be…
15 m/s + 50 m/s = 65 m/s
position-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. Time is a factor twice, making displacement proportional to the square of time.
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 figure this out is through algebra.
For the sake of accuracy, this section should be entitled "One dimensional equations of motion for constant acceleration". Given that such a title would be a stylistic nightmare, let me begin this section with the following qualification. These equations of motion are valid only when acceleration is constant and motion is constrained to a straight line.
Given that we live in a three dimensional universe in which the only constant is change, you may be tempted to dismiss this section outright. It would be correct to say that no object has ever traveled in a straight line and with a constant acceleration anywhere in the universe at any time — not today, not yesterday, not tomorrow, not five billion years ago, not thirty billion years in the future, never. This I can say with absolute
Our goal in this section, is to derive new equations that can be used to describe the motion of an object in terms of its three kinematic variables: velocity (v), position (s), and time (t). There are three ways to pair them up: velocity-time, position-time, and velocity-position. 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.
As long as you are consistent within a problem, it doesn't matter.
velocity-time
The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. 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.
a =
Δv
Δt
Expand Δv to v − v0 and condense Δt to t.
a =
v − v0
t
Then solve for v as a function of t.
v = v0 + at [1]
This is the first equation of motion. It's written like a polynomial — a constant term (v0) followed by a first order term (at). Since the highest order is 1, it's more correct to call it a linear function.
The symbol v0 [vee nought] is called the initial velocity or the velocity a time t = 0. It is often thought of as the "first velocity" but this is a rather naive way to describe it. 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.
The symbol v is the velocity some time t after the initial velocity. It is often called the final velocity but this does not make it an object's "last velocity".
The last part of this equation at is the change in the velocity from the initial value. Recall that ais the rate of change of velocity and that t is the time after some initial event. Rate times time is change. Given an object accelerating at 10 m/s2, after 5 s it would be moving 50 m/s faster. If it started with a velocity of 15 m/s, then its velocity after 5 s would be…
15 m/s + 50 m/s = 65 m/s
position-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. Time is a factor twice, making displacement proportional to the square of time.
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 figure this out is through algebra.
gregheffley:
please tell, it is wrong or right
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The three equations of motion v = u + at ; s = ut + (1/2) at2 and v2 = u2 + 2as can be derived with the help of graphs as described below.1. Derive v = u + at by Graphical MethodConsider the velocity – time graph of a body shown in the below Figure.Velocity–Time graph to derive the equations of motion.The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point Bto OE.Now, Initial velocity of the body, u=OA...... (1)And, Final velocity of the body, v=BC........ (2)But from the graph BC=BD + DCTherefore, v=BD + DC ......... (3)Again DC=OASo, v=BD + OANow, From equation (1), OA=uSo, v=BD + u ........... (4)We should find out the value of BD now. We know that the slope of a velocity – time graph is equal to acceleration, a.Thus, Acceleration, a=slope of line ABor a=BD/ADBut AD=OC = t,so putting t in place of AD in the above relation, we get:a=BD/tor BD=atNow, putting this value of BD in equation (4) we get :v=at + uThis equation can be rearranged to give:v=u + atAnd this is the first equation of motion. It has been derived here by the graphical method.2. Derive s = ut + (1/2) at2 by Graphical MethodVelocity–Time graph to derive the equations of motion.Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus:Distance travelled=Area of figure OABC=Area of rectangle OADC + Area of triangle ABDWe will now find out the area of the rectangle OADCand the area of the triangle ABD.(i) Area of rectangle OADC=OA × OC=u × t=ut ...... (5)(ii) Area of triangle ABD=(1/2) × Area of rectangle AEBD=(1/2) × AD × BD=(1/2) × t × at (because AD = t and BD = at)=(1/2) at2...... (6)So, Distance travelled, s=Area of rectangle OADC + Area of triangle ABDor s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the graphical method.3. Derive v2 = u2 + 2as by Graphical MethodVelocity–Time graph to derive the equations of motion.We have just seen that the distance travelled s by a body in timequatione t is given by the area of the figure OABCwhich is a trapezium. In other words,Distance travelled, s=Area of trapezium OABCNow, OA + CB = u + v and OC = t. Putting these values in the above relation, we get: ...... (7)We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First of motion)
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2as
This is the second equation of motion. It has been derived here by the graphical method.3. Derive v2 = u2 + 2as by Graphical MethodVelocity–Time graph to derive the equations of motion.We have just seen that the distance travelled s by a body in timequatione t is given by the area of the figure OABCwhich is a trapezium. In other words,Distance travelled, s=Area of trapezium OABCNow, OA + CB = u + v and OC = t. Putting these values in the above relation, we get: ...... (7)We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First of motion)
And, at = v – u or Now, putting this value of t in equation (7) above, we get: or 2as=v2 – u2 [because (v + u) × (v – u) = v2 – u2]or v2=u2 + 2as
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