V = Vo + a*t = 31m/s[30o]
Vo - 4*4 = 31*Cos 30 + i31*sin 30
Vo - 16 = 26.8 + 15.5i
Vo = 42.8 + 15.5i
Tan A = Y/X = 15.5/42.8 = 0.36215
A = 19.91o
Vo = 42.8/Cos 19.91 = 45.5m/s[19.9o] =
Initial velocity.
Vo - 4*4 = 31*Cos 30 + i31*sin 30
Vo - 16 = 26.8 + 15.5i
Vo = 42.8 + 15.5i
Tan A = Y/X = 15.5/42.8 = 0.36215
A = 19.91o
Vo = 42.8/Cos 19.91 = 45.5m/s[19.9o] =
Initial velocity.
velocity = initial velocity + acceleration × time
Let's consider the x and y components separately. We have:
velocity_x = initial velocity_x + acceleration_x × time
velocity_y = initial velocity_y + acceleration_y × time
The acceleration is given as 4.0 m/s^2 along the -x axis. So, acceleration_x = -4.0 m/s^2 and acceleration_y = 0.
Since the object is moving at an angle of 30° from the +x axis, the initial velocity vector can be written as:
initial velocity = (initial velocity_x, initial velocity_y)
= (initial velocity * cosθ , initial velocity * sinθ)
Here, θ represents the angle of the initial velocity vector.
We also know that the final velocity (31 m/s) can be written as:
final velocity = (final velocity_x , final velocity_y)
= (velocity * cosθ , velocity * sinθ)
Comparing the x-component of the final and initial velocities, we get:
velocity * cosθ = initial_velocity * cosθ - acceleration_x * time
Similarly, for the y-component:
velocity * sinθ = initial_velocity_y + acceleration_y * time
Substituting the given values, we get:
31 * cos30° = initial_velocity * cos30° - (-4.0) * 4
31 * sin30° = initial_velocity * sin30° + 0 * 4
Simplifying these equations, we have:
31 * √3/2 = initial_velocity * √3/2 + 16
31 * 1/2 = initial_velocity * 1/2
Now, solving for the initial velocity:
initial_velocity = (31 * √3/2 - 16) / √3/2
initial_velocity = (31 * √3 - 32) / √3
So, the initial velocity vector of the object is:
initial velocity = (initial velocity_x, initial velocity_y)
= [(31 * √3 - 32) / √3, (31 * 1/2)]
Thus, the initial velocity vector of the object is approximately:
initial velocity ≈ [39.78 m/s, 15.5 m/s]
vf = vi + at
In this case, you know the final velocity (vf = 31 m/s) and the acceleration (a = -4.0 m/s^2) along the -x axis. However, you don't know the initial velocity (vi) and need to find it.
First, you need to decompose the final velocity vector into its x and y components. To do this, you can use trigonometry. Since the angle is given as 30° counterclockwise from the +x-axis, the angle between the final velocity vector and the +x-axis will be 180° - 30° = 150°.
The x-component of the final velocity can be found using the cosine function:
vfx = vf * cos(theta)
vfx = 31 m/s * cos(150°)
vfx = 31 m/s * (-0.866)
vfx = -26.946 m/s
The y-component of the final velocity can be found using the sine function:
vfy = vf * sin(theta)
vfy = 31 m/s * sin(150°)
vfy = 31 m/s * 0.5
vfy = 15.5 m/s
Now that you have the x and y components of the final velocity, you can substitute them into the equation to find the initial velocity:
vix = vfx - a * t
vix = -26.946 m/s - (-4.0 m/s^2) * 4 s
vix = -26.946 m/s + 16 m/s
vix = -10.946 m/s
viy = vfy
viy = 15.5 m/s
Therefore, the initial velocity vector of the object is vi = (-10.946 m/s, 15.5 m/s) along the -x and +y axes, respectively.