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Prof. Dasmaya Sidhu delivered this lecture at National Institute of Industrial Engineering for Basic Mechanical Engineering course. It includes: Rectilinear, Kinematics, Erratic, Motion, Erratic, Equations, Relationship, Velocity, Slope
Typology: Slides
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a, v, s, t
2
any two of the variables,
a, v, s, t
a = dv/dt,
v = ds/dt,a ds = v dv
Given the
s-t
Graph, construct the
v-t
Graph
•The
s-t
graph can be plotted if the position of the particle
can be
determined experimentally
during a period of time
t.
•To determine the particle’s velocity as a function of time, the
v
-
t
Graph, use
3
the
v
-
t
Graph, use
v = ds/dt
•Velocity as any instant is determined by measuring theslope of the
s-t
graph
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Given the
v-t
Graph, construct the
a-t
Graph
•When the particle’s
v-t
graph is known, the
acceleration as a function of time,
the
a
t
graph can be determined using
5
the
a
t
graph can be determined using
a = dv/dt
•Acceleration as any instant is determined bymeasuring the slope of the v-t graph
a
dt
dv
=
Slope of
v-t
graph = acceleration
6
Given the
a-t
Graph, construct the
v-t
Graph
a-t
graph is known, the
v-t
graph may be
constructed using
a = dv/dt
∫
=
∆
dt
a
v
8
∫
=
∆
dt
a
v
Change in
velocity
Area under
a-t
graph
=
9
Given the
v-t
Graph, construct the
s-t
Graph
v-t
graph is known, the
s-t
graph may
be constructed using
v = ds/dt
11
dt
v
s
Displacement
Area under v-t
graph
=
knowing the initial position
s
0
, and add to this area
increments
s
determined from
v-t
graph
describe each of there segments of the
v-t
graph by
a series of eqns, each of these eqns may be
integrated
to yield
eqns
that describe corresponding segments of
Displacement from v –t graph
12
to yield
eqns
that describe corresponding segments of
the
s-t
graph
Given the
a-s
Graph, construct the
v-s
Graph
v-s
graph can be determined by using
v dv = a ds
integrating this
eqn
between the limit
v = v
at
s =
v-s graph from a –s graph
14
integrating this
eqn
between the limit
v = v
0
at
s =
s
0
and
v = v
1
at
s = s
1
(
)
∫
=
−
1
0
2
0
2
1
1 2
s
s
ds
a
v
v
Area undera-s graph
a-
s
graph
v-s
graph can be obtained from integration, using
vdv = a ds
a-s graph from v –s graph
15
s,v
), the slope
dv/ds
of the
v-s
graph is measured
v
and
dv/ds
are known, the value of
a
can be calculated
a-s graph from v –s graph
17
A bicycle moves along a straight road such that it position isdescribed by the graph as shown.
Construct the
v-t
and
a-t
graphs for 0
≤
t
≤
30s.
Example 12.
18
a-t
Graph.
The
a-t
graph can be determined by differentiating the eqns
defining the lines of the
v-t
graph.
0
6
;
30
10
10
0
=
=
=
≤
<
=
=
=
≤
≤
dt
dv
a
v
s
t
dt
dv
a
t
v
s
t
20
0
6
;
30
10
=
=
=
≤
<
dt
a
v
s
t
The results are plotted.
A test car starts from rest and travels along a straight tracksuch that it accelerates at a constant rate for 10 s and thendecelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t’needed to stop the car. How far has the car traveled?
Example 12.
21
