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
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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](/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
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