Respuesta :
7. The rate at which the velocity of an object changes is the object's rate of acceleration.
8. Each second its speed increases by  8 m/s.
9. The distance the airplane travels each second is 8 m than the distance it traveled during the previous second
10. The ride's rate of acceleration during the  4 seconds is [tex]10 m/s^2[/tex]
Explanation:
7)
The acceleration of an object is defined as the rate of change in velocity of the object. Mathematically:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity of the object
v is the final velocity
t is the time taken for the object's velocity to change from u to v
Acceleration is measured in meters per second squared ([tex]m/s^2[/tex]), and it basically gives a measure of how fast the velocity of an object changes.
Since velocity is a vector quantity, acceleration is a vector quantity as well: this means that an object is accelerated not only when its speed changes, but also when the direction of its velocity changes (for example, in a circular motion).
8)
The velocity of an object moving at constant acceleration is given by
[tex]v=u+at[/tex]
where
u is the initial velocity
v is the velocity at time t
a is the acceleration
We can rewrite it as
[tex]v-u=at[/tex]
where [tex](v-u)[/tex] represents the change in velocity.
For the plane in this problem, its acceleration is
[tex]a=8 m/s^2[/tex]
So, the change in velocity in a time of [tex]t=1 s[/tex] is
[tex](v-u) = at=(8)(1)=8 m/s[/tex]
9)
The distance covered by an object moving at constant acceleration in a time [tex]t[/tex] is
[tex]s=ut+\frac{1}{2}at^2[/tex]
where u is the velocity at the beginning of the time t considered, a is the acceleration, s is the distance covered.
So, the distance covered in a certain second [tex]t_1 = 1 s[/tex] is
[tex]s_1 = u t_1 + \frac{1}{2}at_1^2 = 1u+\frac{1}{2}(8)(1)^2 = u+4[/tex]
After that second, the new velocity is
[tex]u'=u+at_1 = u+(8)(1) = u+8[/tex]
The distance travelled in the next second is
[tex]s_2 = u' t_2 + \frac{1}{2}at^2 = (u+8)(1) + \frac{1}{2}(8)(1)^2=u+8+4=u+12[/tex]
So, the difference in the distance travelled in the next second and in the previous second is
[tex]s_2-s_1 = (u+12)-(u+4)=8 m[/tex]
10)
The acceleration of the ride is given by
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time interval
Here we have
u = 0 m/s
v = 40 m/s
t = 4 s
Substituting,
[tex]a=\frac{40-0}{4}=10 m/s^2[/tex]
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