Acceleration of a object 4m/s^2 then average speed 2s between 6s
Answers
Explanation:
We're asked to find the average speed of a particle during a motion.
The equation for average speed is
v
av
=
distance traveled
Δ
t
where
Δ
t
is the time interval
We know it uniformly accelerated from rest at a rate of
a
x
=
4
m/s
2
for
t
=
4
s
We can use the kinematics equation
Δ
x
=
v
0
x
t
+
1
2
a
x
t
2
−−−−−−−−−−−−−−−−−
to find the distance it travels
Δ
x
during this acceleration.
Here, since it started from rest, the initial velocity
v
0
x
is
0
, leaving us with
Δ
x
=
1
2
a
x
t
2
−−−−−−−−−−−
Plugging in known values:
Δ
x
1
=
1
2
(
4
l
m/s
2
)
(
4
l
s
)
2
=
32
l
m
−−−−
For the second part of this motion, we're given that the acceleration is constant (and unknown) and that it comes to rest in
8
s
.
We need to find the velocity of the particle after it has finished its first acceleration, using the equation
Δ
x
1
=
(
v
0
x
+
v
x
2
)
t
where
v
x
is the velocity in question, and
v
0
x
is still
0
as before:
32
l
m
=
(
0
+
v
x
2
)
(
4
l
s
)
v
x
=
16
l
m/s
−−−−−−
(you could've also used the equation
v
x
=
v
0
x
+
a
x
t
to find the velocity.)
This value represents the initial velocity of the particle as it begins to negatively accelerate. We can now use the same equation
Δ
x
2
=
(
v
0
x
+
v
x
2
)
t
to find the distance traveled
Δ
x
.
Here,
v
x
, the final velocity, is
0
(it comes to rest)
v
0
x
is
16
l
m/s
t
is
8
s
:
Δ
x
2
=
(
16
l
m/s
+
0
2
)
(
8
l
s
)
=
64
l
m
−−−−
The total distance traveled is
distance traveled
=
Δ
x
1
+
Δ
x
2
=
32
l
m
+
64
l
m
=
96
l
m
−−−−
And the time
Δ
t
is
Δ
t
=
4
s
+
8
s
=
12
l
s
−−−−
Thus, the average speed of the particle is
v
av
=
96
l
m
12
l
s
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
8
l
m/s
∣
∣
−−−−−−−−−−−