comes from our This equation, which of of constant acceleration can be put in yet another form which does not depend so much Before going to examples lets show first that our three equations for the special case on when we started our clock or where we called x = 0. For example at t = t2 we have handy equations for the special case, will be very useful. 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(t₂) = a (t₂-t₁) +v(t₁). This really says the same thing as the original equation: The velocity at the end of a time interval is the velocity at the beginning plus the acceleration times the time interval. Doing the same thing for the second equation we have v(t₁) = at₁ +v(0). 1 x(t₂) = at +v(0)t₂ + x(0) x(t₁) = = (t₂) - (t₁) = at +v(0)t₁ + x(0) 1 2a. (t – tỉ) + n(0)(tz - t) Exercise: Go through the algebra to show that the above leads to 1 x(t₂) − x(t₁) = a · (t₂ − t₁)² + v(t₁)(t2 − t₁). HINT: You have to use v(t₁) = at₁ +v(0). (This equation also has the same interpretation as the original equation if one considers time intervals.) Exercise: Show that the third formula, relating the values of v², leads to v² (1₂) - v² (t₁) = 2a [x(t₂) - (t₁)]. 1
comes from our This equation, which of of constant acceleration can be put in yet another form which does not depend so much Before going to examples lets show first that our three equations for the special case on when we started our clock or where we called x = 0. For example at t = t2 we have handy equations for the special case, will be very useful. 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(t₂) = a (t₂-t₁) +v(t₁). This really says the same thing as the original equation: The velocity at the end of a time interval is the velocity at the beginning plus the acceleration times the time interval. Doing the same thing for the second equation we have v(t₁) = at₁ +v(0). 1 x(t₂) = at +v(0)t₂ + x(0) x(t₁) = = (t₂) - (t₁) = at +v(0)t₁ + x(0) 1 2a. (t – tỉ) + n(0)(tz - t) Exercise: Go through the algebra to show that the above leads to 1 x(t₂) − x(t₁) = a · (t₂ − t₁)² + v(t₁)(t2 − t₁). HINT: You have to use v(t₁) = at₁ +v(0). (This equation also has the same interpretation as the original equation if one considers time intervals.) Exercise: Show that the third formula, relating the values of v², leads to v² (1₂) - v² (t₁) = 2a [x(t₂) - (t₁)]. 1
University Physics Volume 3
17th Edition
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:William Moebs, Jeff Sanny
Chapter5: Relativity
Section: Chapter Questions
Problem 88AP: A spacecraft starts from being at rest at the origin and accelerates at a constant rate g, as seen...
Related questions
Concept explainers
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
Topic Video
Question
100%
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd0ee4640-6745-4736-a864-91941cf0214c%2F6973e20b-0a59-42c1-b037-7cd669561b90%2Fxjrom5_processed.jpeg&w=3840&q=75)
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
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
University Physics Volume 3
Physics
ISBN:
9781938168185
Author:
William Moebs, Jeff Sanny
Publisher:
OpenStax
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Classical Dynamics of Particles and Systems
Physics
ISBN:
9780534408961
Author:
Stephen T. Thornton, Jerry B. Marion
Publisher:
Cengage Learning
University Physics Volume 3
Physics
ISBN:
9781938168185
Author:
William Moebs, Jeff Sanny
Publisher:
OpenStax
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Classical Dynamics of Particles and Systems
Physics
ISBN:
9780534408961
Author:
Stephen T. Thornton, Jerry B. Marion
Publisher:
Cengage Learning
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning