Answer:
Distance traveled
s
=
1721.344
m
or
≈
1.721
k
m
Explanation:
Aide memoires
There are three equations for 1D motion involving constant acceleration.
v
=
u
+
a
t
s
=
u
t
+
1
2
a
t
2
v
2
−
u
2
=
2
a
s
where symbols have meanings ascribed to each.
In the problem given quantities are acceleration
a
, time
t
. And we need to find distance
s
traveled.
Inspection reveals that equation 2 is the most suitable as it has all three quantities of interest in it except initial velocity
u
.
Answer starts from here
→
We know that
s
=
u
t
+
1
2
a
t
2
where
s
is the distance traveled,
a
is constant acceleration experienced by an object,
u
is its initial velocity and
t
time of travel under this acceleration.
Now we assume that the airplane starts from rest on the runway. Therefore, first term on the RHS is zero. Inserting given values in the equation
s
=
1
2
×
3.20
×
32.8
2
s
=
1721.344
m
