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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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On the excercise in the image, I am unable to figure out how to solve these algebraically.
![This equation, which of cou
handy equations for the special case, will be very useful.
comes from c
of constant acceleration can be put in yet another form which does not depend
Before going to examples lets show first that our three equations for the special
on when we started our clock or where we called x = 0. For example at t = t2 we have
v(t₂) = at₂ + v(0)
or
and at t₁ we have
Subtracting the second from the first gives
and subtracting,
v(t₂) - v(t₁) = a (t₂ - 1₁)
v(t₁) = at₁ +v(0).
v(t₂) = a (t₂-t₁) +v(t₁).
interval is the velocity at the beginning plus the acceleration times the time interval.
Doing the same thing for the second equation we have
This really says the same thing as the original equation: The velocity at the end of a time
x (t₂) =
z(t₁) =
x(t₂) - x(t₁) =
1
at +v(0)t₂ + x(0)
at² + v(0)t₁ + x(0)
1
·
2
x(1₂) − x(t₁) = a (t²- t²) + v(0)(t₂ − t₁)
-
Exercise: Go through the algebra to show that the above leads to noitsups broosa edT
case
17 a
a. (t₂-t₁)² + v(t₁)(t2 — t₁).
2
as the original equation if one considers time intervals.)
HINT: You have to use v(t₁) = at₁ +v(0). (This equation also has the same interpretation
Exercise: Show that the third formula, relating the values of v², leads to
v²(t₂) - v²(t₁) = 2a[r(t₂) - x(t₁)].
22
st
i](https://content.bartleby.com/qna-images/question/d0ee4640-6745-4736-a864-91941cf0214c/6973e20b-0a59-42c1-b037-7cd669561b90/xjrom5_thumbnail.jpeg)
Transcribed Image Text:This equation, which of cou
handy equations for the special case, will be very useful.
comes from c
of constant acceleration can be put in yet another form which does not depend
Before going to examples lets show first that our three equations for the special
on when we started our clock or where we called x = 0. For example at t = t2 we have
v(t₂) = at₂ + v(0)
or
and at t₁ we have
Subtracting the second from the first gives
and subtracting,
v(t₂) - v(t₁) = a (t₂ - 1₁)
v(t₁) = at₁ +v(0).
v(t₂) = a (t₂-t₁) +v(t₁).
interval is the velocity at the beginning plus the acceleration times the time interval.
Doing the same thing for the second equation we have
This really says the same thing as the original equation: The velocity at the end of a time
x (t₂) =
z(t₁) =
x(t₂) - x(t₁) =
1
at +v(0)t₂ + x(0)
at² + v(0)t₁ + x(0)
1
·
2
x(1₂) − x(t₁) = a (t²- t²) + v(0)(t₂ − t₁)
-
Exercise: Go through the algebra to show that the above leads to noitsups broosa edT
case
17 a
a. (t₂-t₁)² + v(t₁)(t2 — t₁).
2
as the original equation if one considers time intervals.)
HINT: You have to use v(t₁) = at₁ +v(0). (This equation also has the same interpretation
Exercise: Show that the third formula, relating the values of v², leads to
v²(t₂) - v²(t₁) = 2a[r(t₂) - x(t₁)].
22
st
i
Expert Solution
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- Let C be the region between the circles x² + y² = 1 and x² + y² = 4. Let 2 2 2 F(x, y) = (²3³-²3V², 30³² Compute the circulation across C. Flux = y³, x³ - 3x²arrow_forwardSolve for #3arrow_forwardCan you can explain the proof of "the equation of continuity", in an easy-to-understand manner? Attached images are the proof my professor gave. I really do not understand all the simple v's , capital v's and all? Can you kindly explain it in an easy-to-understand manner? Thank you!arrow_forward
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