Answer :
To find the displacement of the plane, we can use the formula for displacement when average velocity and time are known:
[tex]\[
\Delta x = v \times t
\][/tex]
Where:
- [tex]\(\Delta x\)[/tex] is the displacement.
- [tex]\(v\)[/tex] is the average velocity.
- [tex]\(t\)[/tex] is the time.
In this problem, the average velocity [tex]\(v\)[/tex] is [tex]\(-98.5 \, \text{m/s}\)[/tex] and the time [tex]\(t\)[/tex] is [tex]\(45.0 \, \text{s}\)[/tex].
Now, let's calculate the displacement:
[tex]\[
\Delta x = (-98.5 \, \text{m/s}) \times (45.0 \, \text{s})
\][/tex]
[tex]\[
\Delta x = -4432.5 \, \text{m}
\][/tex]
So, the displacement of the plane is [tex]\(-4432.5\)[/tex] meters. The negative sign indicates that the displacement is in the opposite direction of the positive reference direction.
[tex]\[
\Delta x = v \times t
\][/tex]
Where:
- [tex]\(\Delta x\)[/tex] is the displacement.
- [tex]\(v\)[/tex] is the average velocity.
- [tex]\(t\)[/tex] is the time.
In this problem, the average velocity [tex]\(v\)[/tex] is [tex]\(-98.5 \, \text{m/s}\)[/tex] and the time [tex]\(t\)[/tex] is [tex]\(45.0 \, \text{s}\)[/tex].
Now, let's calculate the displacement:
[tex]\[
\Delta x = (-98.5 \, \text{m/s}) \times (45.0 \, \text{s})
\][/tex]
[tex]\[
\Delta x = -4432.5 \, \text{m}
\][/tex]
So, the displacement of the plane is [tex]\(-4432.5\)[/tex] meters. The negative sign indicates that the displacement is in the opposite direction of the positive reference direction.
