A) Acceleration: ![a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2](https://tex.z-dn.net/?f=a_1%20%3D%200.75%20m%2Fs%5E2%2C%20a_2%20%3D%200%2C%20a_3%20%3D%20-1.57%20m%2Fs%5E2)
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:
![a=\frac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7D%7Bt%7D)
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
(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: ![a_2 = 0](https://tex.z-dn.net/?f=a_2%20%3D%200)
- In the third segment,
u = 11 m/s (north)
v = 0 (he comes to a stop)
t = 7 s
So the acceleration is
![a=\frac{0-11}{7}=-1.57 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B0-11%7D%7B7%7D%3D-1.57%20m%2Fs%5E2)
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:
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
u is the initial velocity
a is the acceleration
t is the time
- For the first segment, we have
![u = 0\\a = 0.75 m/s^2\\t=4 s](https://tex.z-dn.net/?f=u%20%3D%200%5C%5Ca%20%3D%200.75%20m%2Fs%5E2%5C%5Ct%3D4%20s)
So the displacement is
![s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m](https://tex.z-dn.net/?f=s_1%20%3D%200%2B%5Cfrac%7B1%7D%7B2%7D%280.75%29%284%29%5E2%3D6%20m)
- For the second segment, we have
![u = 11 m/s\\a = 0\\t=15 s](https://tex.z-dn.net/?f=u%20%3D%2011%20m%2Fs%5C%5Ca%20%3D%200%5C%5Ct%3D15%20s)
So the displacement is
![s_2 = (11)(15)+0=165 m](https://tex.z-dn.net/?f=s_2%20%3D%20%2811%29%2815%29%2B0%3D165%20m)
- For the third segment, we have
![u = 11\\a = -1.57 m/s^2\\t=7 s](https://tex.z-dn.net/?f=u%20%3D%2011%5C%5Ca%20%3D%20-1.57%20m%2Fs%5E2%5C%5Ct%3D7%20s)
So the displacement is
![s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m](https://tex.z-dn.net/?f=s_3%20%3D%20%2811%29%287%29%2B%5Cfrac%7B1%7D%7B2%7D%28-1.57%29%287%29%5E2%3D38.5%20m)
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:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
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
(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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