1) Time required to stop: 16.3 s
2) Distance required to stop the plane: 0.91 km
Explanation:
1)
The motion of the plane is a uniformly accelerated motion, therefore we can use the following suvat equation:
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
When the plane stops, we have:
v = 0
u = 111 m/s (initial velocity)
[tex]a=-6.8 m/s^2[/tex] (acceleration)
Therefore, the time the plane takes to stop is:
[tex]t=\frac{v-u}{a}=\frac{0-111.1}{-6.8}=16.3 s[/tex]
2)
To find the distance required for the plane to stop, we can use another suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the stopping distance
Since the plane stops,
v = 0
u = 111.1 m/s
[tex]a=-6.8 m/s^2[/tex]
Therefore, the distance covered by the plane while stopping is:
[tex]s=\frac{v^2-u^2}{2a}=\frac{(0)^2-(111.1)^2}{2(-6.8)}=908 m = 0.91 km[/tex]
Learn more about accelerated motion:
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