Respuesta :
Answer:
e.26m/s
Explanation:
Vf=Vi+at (1)
Vf=9j+(2i-4j)t
X= X₀+at
now, in the i direction
15=O+2t or t=7.5 when x position is 15
Lets put that into the (1) equation, solve for Vf.
Vf=9j+(2i-4j)7.5
Vf= 15i - 21j
Speed= [tex]\sqrt{xcomponent^2 + ycomponent^2}[/tex]
Vf= 25.8 m/s
The speed of the particle when it is at x = 15m after leaving the origin with a velocity of 9.0 m/s in the y-direction and moving in the xy plane with a constant acceleration of (2.0i - 4.0j) m/s² is 10 m/s (option a).
We can find the speed of the particle with the following equation:
[tex] v_{f}^{2} = v_{i}^{2} + 2ad [/tex]
Where:
- [tex] v_{f} [/tex]: is the final velocity
- [tex] v_{i} [/tex]: is the initial velocity
- a: is the acceleration
- d: is the distance
The magnitude of the final speed is given by:
[tex] v_{f} = \sqrt{v_{f}_{x}^{2} + v_{f}_{y}^{2}} [/tex] (1)
So, we need to find the horizontal (x) and vertical (y) components of the velocity.
The horizontal component
[tex] v_{f}_{x}^{2} = v_{i}_{x}^{2} + 2a_{x}x [/tex] (2)
Where:
- x: is the horizontal position = 15 m
- [tex]a_{x}[/tex]: is the horizontal component of the acceleration = 2.0 m/s²
- [tex]v_{i}_{x}[/tex]: is the horizontal component of the initial velocity = 0
Then, the x-component of v is (eq 2):
[tex] v_{f}_{x} = \sqrt{2*2.0 m/s^{2}*15 m} = 7.75 m/s [/tex]
The vertical component
[tex] v_{f}_{y}^{2} = v_{i}_{y}^{2} + 2a_{y}y [/tex] (3)
Where:
- y: is the vertical position
- [tex]a_{y}[/tex]: is the vertical component of the acceleration = -4.0 m/s²
- [tex]v_{i}_{y}[/tex]: is the vertical component of the initial velocity = 9.0 m/s
We can find the vertical position with the time:
[tex] x_{f} = x_{i} + v_{i}_{x}t + \frac{1}{2}a_{x}t^{2} [/tex]
[tex] 15 m = 0 + 0*t + \frac{1}{2}2.0 m/s^{2}*t^{2} [/tex]
[tex] t = \sqrt{\frac{2*15m}{2.0 m/s^{2}}} = 3.87 s [/tex]
Now, the vertical position is:
[tex] y_{f} = y_{i} + v_{i}_{y}t + \frac{1}{2}a_{y}t^{2} [/tex]
[tex] y_{f} = 0 + 9.0 m/s*3.87s + \frac{1}{2}(-4.0 m/s^{2})*(3.87 s)^{2} = 4.88 m [/tex]
Hence, the vertical component of the velocity is (eq 3):
[tex] v_{f}_{y} = \sqrt{(9.0 m/s)^{2} + 2*(-4.0 m/s^{2})*4.88 m} = 6.48 m/s [/tex]
Finally, the magnitude of the final speed is (eq 1):
[tex] v_{f} = \sqrt{(7.75 m/s)^{2} + (6.48 m/s)^{2}} = 10.10 [/tex]
Therefore, the speed of the particle is 10 m/s (option a).
Learn more about speed here:
- https://brainly.com/question/7667446?referrer=searchResults
- https://brainly.com/question/936572?referrer=searchResults
I hope it helps you!
