1. Resolve an acceleration vector of 40m/s^2 [25 degrees E of S] into its components
For this I got 17 as my x component and 36 as my y component
2. A player is positioned 35m [40 degrees W of S] of the net. He shoots a puoock 25m [E] to a teammate. What second displacement does the puck have to travel in order to make it to the net?
My answer for this is 23m [S 45 degrees E]
3. Resolve a displacement vector of 5 m [E] into its components.
My answer for the x component is 5m and the y component is 0
#1 ok
#2 I get that the player is at (-22.5,-26.8) from the net, so his shot takes the puck to (2.5,-26.8). Making it 26.9 from the net. You can figure the angle.
How did you get your solution?
#3 ok
1. X = 40*cos25 = 36.21 m/s^2.
Y = 40*sin25 = 16.90 m/s^2.
2. d2-d1 = 25 m.
d2 = 25 + d1
d2=25 + 35[230o]=25+35*cos230+i35*sin230
= 25 - 22.50-26.81i = 2.50 - 26.81i =
27m[-84.7o] = 27m[275.3o] = 27m[5.3o E.
of S.]
3. Correct.
1. To resolve the acceleration vector into its components, you can use trigonometry. Start by drawing a diagram representing the vector. In this case, the vector is 40 m/s^2 and it is 25 degrees east of south.
To find the x-component, you need to find the adjacent side of the triangle formed by the vector. Use the cosine function:
cos(25) = adjacent/hypotenuse
cos(25) = x-component/40
Rearranging the equation, we get:
x-component = 40 * cos(25)
Evaluating this expression, we find that the x-component is approximately 35.056 m/s^2.
To find the y-component, you need to find the opposite side of the triangle formed by the vector. Use the sine function:
sin(25) = opposite/hypotenuse
sin(25) = y-component/40
Rearranging the equation, we get:
y-component = 40 * sin(25)
Evaluating this expression, we find that the y-component is approximately 16.717 m/s^2.
Therefore, the x-component is approximately 35.056 m/s^2 and the y-component is approximately 16.717 m/s^2.
2. To find the second displacement of the puck, we need to calculate the resultant displacement vector. The player is positioned 35 m west of the net, which gives us the x-component of the displacement. The puck is shot 25 m east, which gives us the x-component of the second displacement. We need to calculate the y-component of the second displacement.
To find the y-component, we can use the tangent function:
tan(45) = y-component/second displacement
Rearranging the equation, we get:
y-component = second displacement * tan(45)
We know that the x-component of the first displacement is 35 m and the x-component of the second displacement is 25 m. Assuming the second displacement is in the southeast direction, we can add the x-components and subtract the y-components.
x-component: 35 m - 25 m = 10 m (west)
y-component: second displacement * tan(45) = second displacement
Therefore, the second displacement has to be equal to the y-component, which is approximately 23 m, in the direction 45 degrees east of south.
3. Since the displacement vector is in the east direction and there is no displacement in the y-direction, the x-component is simply equal to the magnitude of the displacement, which is 5 m. The y-component is 0 since there is no displacement in that direction.
Therefore, the x-component is 5 m and the y-component is 0.