draw a sectional view of human heart and label the following parts .A. aorta .B. pulmonary veins.C. Vena cava .D. left ventricle
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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 since velocity is now 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 were so simple. This example only works when initial velocity is zero. 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. 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. 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. One way to figure them out is to use algebra.
Start with the definition of velocity.
v̅ = ΔsΔt
Expand Δs to s − s0 and condense Δt to t.
v̅ = s − s0t
Solve for position.
s = s0 + v̅t [a]
To continue, we need to resort to a little trick known as the mean speed theorem or the Merton rule. I prefer the latter since the rule can be applied to any quantity that changes at a uniform rate — not just speed. The Merton rule was first published in 1335 at Merton College, Oxford by the English philosopher, mathematician, logician, and calculator WilliamHeytesbury. When the rate of change of a quantity is constant, the quantity changes at a uniform rate so that its average value is halfway between its initial and final values.
v̅ = ½(v + v0) [4]
Substitute the first equation of motion [1] into this equation [4] and simplify with the intent of eliminating v.
v̅ = ½[(v0 + at) + v0] v̅ = ½(2v0 + at) v̅ = v0 + ½at [b]
Now substitute [b] into [a] to eliminate v̅ [vee bar].
s = s0 + (v0 + ½at)t
And finally, solve for s as a function of t.
s = s0 + v0t + ½at2 [2]
This is the second equation of motion. It's written like a polynomial — a constant term (s0), followed by a first order term (v0t ), followed by a second order term (½at2). Since the highest order is 2, it's more correct to call it a quadratic.
Would that it were so simple. This example only works when initial velocity is zero. 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. 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. 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. One way to figure them out is to use algebra.
Start with the definition of velocity.
v̅ = ΔsΔt
Expand Δs to s − s0 and condense Δt to t.
v̅ = s − s0t
Solve for position.
s = s0 + v̅t [a]
To continue, we need to resort to a little trick known as the mean speed theorem or the Merton rule. I prefer the latter since the rule can be applied to any quantity that changes at a uniform rate — not just speed. The Merton rule was first published in 1335 at Merton College, Oxford by the English philosopher, mathematician, logician, and calculator WilliamHeytesbury. When the rate of change of a quantity is constant, the quantity changes at a uniform rate so that its average value is halfway between its initial and final values.
v̅ = ½(v + v0) [4]
Substitute the first equation of motion [1] into this equation [4] and simplify with the intent of eliminating v.
v̅ = ½[(v0 + at) + v0] v̅ = ½(2v0 + at) v̅ = v0 + ½at [b]
Now substitute [b] into [a] to eliminate v̅ [vee bar].
s = s0 + (v0 + ½at)t
And finally, solve for s as a function of t.
s = s0 + v0t + ½at2 [2]
This is the second equation of motion. It's written like a polynomial — a constant term (s0), followed by a first order term (v0t ), followed by a second order term (½at2). Since the highest order is 2, it's more correct to call it a quadratic.
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