A) Acceleration: [tex]a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2[/tex]
B) The total displacement is 209.5 m north
C) The average velocity is 8.06 m/s north
Explanation:
A)
Acceleration is defined as:
[tex]a=\frac{v-u}{t}[/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken for the velocity to change from u to v
Here we have:
- In the first segment,
u = 8 m/s north
v = 11 m/s north
t = 4 s
So the acceleration is
[tex]a_1 = \frac{11-8}{4}=0.75 m/s^2[/tex] (north)
- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,
u = v
Therefore, the acceleration is zero: [tex]a_2 = 0[/tex]
- In the third segment,
u = 11 m/s (north)
v = 0 (he comes to a stop)
t = 7 s
So the acceleration is
[tex]a=\frac{0-11}{7}=-1.57 m/s^2[/tex]
And the negative sign means the acceleration is south, opposite to the direction of motion.
B)
In a uniformly accelerated motion, the displacement can be calculated as:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u is the initial velocity
a is the acceleration
t is the time
- For the first segment, we have
[tex]u = 0\\a = 0.75 m/s^2\\t=4 s[/tex]
So the displacement is
[tex]s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m[/tex]
- For the second segment, we have
[tex]u = 11 m/s\\a = 0\\t=15 s[/tex]
So the displacement is
[tex]s_2 = (11)(15)+0=165 m[/tex]
- For the third segment, we have
[tex]u = 11\\a = -1.57 m/s^2\\t=7 s[/tex]
So the displacement is
[tex]s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m[/tex]
So the total displacement is:
s = 6 m + 165 m + 38.5 m = 209.5 m
In the north direction (positive direction)
C)
The average velocity is given by:
[tex]v=\frac{d}{t}[/tex]
where
d is the total displacement
t is the total time
Here we have:
d = 209.5 m
t = 26 s
Therefore, the average velocity is
[tex]v=\frac{209.5}{26}=8.06 m/s[/tex] (north)
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
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