6a) Vector: 17.0 m/s @
x-component:
y-component:
6b) Vector: 13.0 m/s^2 @
x-component:
y-component:
6c) Vector: 19.0 m @
x-component:
y-component:
6d) Vector: 22.0 m/s @
x-component:
y-component:
7a) Components: Ax =34.0m,Ay =-34.0m
Magnitude:
Direction:
7b) Components: Bx =0.00km,By =150.0km
Magnitude:
Direction: since it is in the y-direction (no component on x)
7c) Components: Cx = -8.00 m/s2, Cy = 6.00 m/s2
Magnitude:
Direction:
8) In this problem, 0.750 m corresponds to the vertical side and 1.25 m corresponds to the horizontal side of a right triangle. The length of the stick corresponds to the length of the hypothenuse of this triangle, that can be found using the Pytagorean's theorem:
The angle the stick makes with the ground is given by:
9) The displacement of the boat corresponds to a vector of length L=300 km and angle with respect to east. Therefore, the two components in the north and east directions are:
- north:
- east:
10) The two accelerations correspond to the two sides of a right triangle, therefore the resultant acceleration corresponds to the length of the hypothenuse of the triangle:
11a) [email protected] [email protected]
Let's resolve each vector in its components:
Now we sum the components in each direction:
So the magnitude of the resultant vector is:
And the direction is:
11b) 22.0 m/s @ 225o + 40.0 m/s @ 315o
Let's resolve each vector in its components:
Now we sum the components in each direction:
So the magnitude of the resultant vector is:
And the direction is:
11c) 17.0 m/s2 @ 330o - 11.0 m/s2 @ 270o
Let's resolve each vector in its components:
Now we calculate the difference between the components in each direction:
So the magnitude of the resultant vector is:
And the direction is:
12a) The change in velocity is equal to the vector difference between the final velocity (vf) and the initial velocity (vi), so let's proceed as in the previous exercise:
Difference:
Magnitude of the resultant vector:
12b) Acceleration: