) At an instant, the horizontal position of a particle is given by x = 8t, where xis in m and t is in sec. If the equation of path is given by y = x / 10 at t = 2 sec. Find: i) Magnitude of velocity. ii) Magnitude of acceleration
Answers
Answer:
ok
Explanation:
Taking the derivative with respect to time
→
v
(
t
)
,
we find
→
a
(
t
)
=
lim
t
→
0
→
v
(
t
+
Δ
t
)
−
→
v
(
t
)
Δ
t
=
d
→
v
(
t
)
d
t
.
The acceleration in terms of components is
→
a
(
t
)
=
d
v
x
(
t
)
d
t
^
i
+
d
v
y
(
t
)
d
t
^
j
+
d
v
z
(
t
)
d
t
^
k
.
Also, since the velocity is the derivative of the position function, we can write the acceleration in terms of the second derivative of the position function:
→
a
(
t
)
=
d
2
x
(
t
)
d
t
2
^
i
+
d
2
y
(
t
)
d
t
2
^
j
+
d
2
z
(
t
)
d
t
2
^
k
.
EXAMPLE
Finding an Acceleration Vector
A particle has a velocity of
→
v
(
t
)
=
5.0
t
^
i
+
t
2
^
j
−
2.0
t
3
^
k
m/s
.
(a) What is the acceleration function? (b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.
Solution
Show Answer
Significance
In this example we find that acceleration has a time dependence and is changing throughout the motion. Let’s consider a different velocity function for the particle.
EXAMPLE
Finding a Particle Acceleration
A particle has a position function
→
r
(
t
)
=
(
10
t
−
t
2
)
^
i
+
5
t
^
j
+
5
t
^
k
m
.
(a) What is the velocity? (b) What is the acceleration? (c) Describe the motion from t = 0 s.2.method is
A similar set of kinematic equations could be written for motion in the z-direction:
x
(
t
)
=
x
0
+
(
v
x
)
avg
t
v
x
(
t
)
=
v
0
x
+
a
x
t
x
(
t
)
=
x
0
+
v
0
x
t
+
1
2
a
x
t
2
v
2
x
(
t
)
=
v
2
0
x
+
2
a
x
(
x
−
x
0
)
y
(
t
)
=
y
0
+
(
v
y
)
avg
t
v
y
(
t
)
=
v
0
y
+
a
y
t
y
(
t
)
=
y
0
+
v
0
y
t
+
1
2
a
y
t
2
v
2
y
(
t
)
=
v
2
0
y
+
2
a
y
(
y
−
y
0
)
.