Answer:
V= 13 m/s
Explanation:
We're asked to find the magnitude of the object's velocity after a force is applied for
3
s
parallel to its motion.
We'll call the direction it's moving the positive
x
-axis, and the direction of the applied force the positive
y
-axis.
The components of the initial velocity are
v
0
x
=
5
m/s
v
0
y
=
0
(It's moving at
5
meters per second in the straight line we called the
x
-axis.)
We know the object's mass is
3
kg
, and the force applied is
12
N
in the positive
y
-direction. The magnitude of the constant acceleration is thus
a
y
=
∑
F
y
m
=
12
l
N
3
l
kg
=
4
m/s
2
Since this acceleration is directed upward, and the initial
y
-velocity is
0
, we can use the kinematics equation
v
y
=
v
0
y
+
a
y
t
to find the
y
-velocity after
3
seconds.
Plugging in known values, we have
v
y
=
0
+
(
4
l
m/s
2
)
(
3
l
s
)
=
12
m/s
No acceleration was applied in the
x
-direction, so it's
x
-velocity remains
5
m/s
. The magnitude of the velocity is thus
v
=
√
(
v
x
)
2
+
(
v
y
)
2
=
√
(
5
l
m/s
)
2
+
(
12
l
m/s
)
2
=
13
m/s
