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Byju's Answer
Standard XII
Physics
Distance
Is average ve...
Question
Is average velocity ; total displacement /total time (or) x
2
-x
1
/ t
2
-t
1
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Solution
Yes, it is equal to total displacement divided by total time.
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1
Similar questions
Q.
What is the difference between these two cases:
(Please explain me in simple words)
1) Average velocity = Total Displacement / Total time
2) Average velocity = (Initial velocity + Final velocity) / 2
= [(u+v)/2]
Q.
Starting from rest a particle is first accelerated for time
t
1
with constant acceleration
a
1
and then stops in time
t
2
with constant retardation
a
2
.
Let
v
1
be the average velocity in this case and
s
1
the total displacement. In the second case
it is accelerated for the same time
t
1
with constant acceleration
2
a
1
and come to rest with constant retardation
a
2
in time
t
3
.
If
v
2
is the average velocity in this case and
s
2
the total displacement, then
Q.
A particle is moving a long
x
−
a
x
i
s
at time
t
1
=
2
s
, its position is
x
1
=
3
m
and at time
t
2
=
7
s
, its position is
x
2
=
18
m
.The average velocity of the particle is
Q.
Starting form rest a particle is first accelerated for time
t
1
with constant acceleration
a
1
and then stops in time
t
2
with constant retardation
a
2
. Let
v
1
be the average velocity in this case and
s
1
be the total displacement. In the second case, it is accelerated for the same time
t
1
with constant acceleration
2
a
1
and comes to rest with constant retardation
a
2
in time
t
3
. If
v
2
is the average velocity in this case and
s
2
the total displacement. Then
Q.
The area under the velocity-time graph between any two instants
t
=
t
1
and
t
=
t
2
gives the displacement in time
δ
t
=
t
2
−
t
1
:
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