a) See graph in attachment
b) The suvat equation to use is [tex]v_f^2 - v_i^2 = 2as[/tex]
c) The acceleration is [tex]a=\frac{v_f^2-v_i^2}{2s}[/tex]
d) The acceleration is [tex]1.25 m/s^2[/tex]
e) The time needed is 8 s
Explanation:
a)
For this part, find in attachment the diagram representing this situation.
Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.
Then we have the following:
- The initial position of the boat is [tex]x_i = 0[/tex], the origin
- The final position of the boat is [tex]x_f = 200 m[/tex]
- The initial velocity of the boat is [tex]v_i = 20.0 m/s[/tex]
- The final velocity of the boat is [tex]v_f = 30.0 m/s[/tex]
Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.
b)
The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:
[tex]v_i = 20.0 m/s[/tex], the initial velocity
[tex]v_f = 30.0 m/s[/tex], the final velocity
[tex]s = x_f - x_i = 200 m[/tex], the displacement of the boat
Therefore, the equation that best can be use to find the acceleration is
[tex]v_f^2 - v_i^2 = 2as[/tex]
where
[tex]a[/tex] is the acceleration
c)
Now we have to solve the equation
[tex]v_f^2 - v_i^2 = 2as[/tex]
In order to find the acceleration.
This can be done by dividing both terms by [tex]2s[/tex]: this way, we find
[tex]\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}[/tex]
And so the acceleration is
[tex]a=\frac{v_f^2-v_i^2}{2s}[/tex]
d)
Now we can use the equation found in part c) in order to find the acceleration.
We have the following data:
[tex]v_i = 20.0 m/s[/tex], the initial velocity
[tex]v_f = 30.0 m/s[/tex], the final velocity
[tex]s = x_f - x_i = 200 m[/tex], the displacement of the boat
And substituting into the equation,
[tex]a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2[/tex]
e)
In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:
[tex]v_f = v_i + at[/tex]
where:
[tex]v_i[/tex] is the initial velocity
[tex]v_f[/tex] is the final velocity
a is the acceleration
t is the time
Here we have:
[tex]v_i = 20.0 m/s[/tex]
[tex]v_f = 30.0 m/s[/tex]
[tex]a=1.25 m/s^2[/tex]
Solving for t, we find:
[tex]t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s[/tex]
Learn more about accelerated motion:
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