Ryan runs 50 m north and then 50 m east. What is his displacement?
1 answer:
You might be interested in
The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude
the force of the wind F, acting horizontally, with intensity
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
By dividing the second equation by the first one, we get
From which we find
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
We have: the weight, downward, with magnitude
the force of the wind F, acting horizontally, with intensity
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
By dividing the second equation by the first one, we get
From which we find
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
1) Analyze the equation
Vₓ = α + β t²
Vₓ = 3.00 + 0.100 t²
That is a quadratic equation, so the graph is a parabola.
2) Therefore, take into account that the main points to draw a parabola are:
i) vertex
ii) concavity
iii) y-intercepts
iv) x - intercepts
Also, for all graphs you need the domain and the range.
3) Find the y-intercept (t = 0)
t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00
4) Find the x-intercepts (Vₓ = 0)
Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²
⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.
5) Concavity
Since, the coefficient of t² is positive, the parabola open upwards.
6) Vertex
It is the local minimum of the equation. You can find it by the first derivative
Vₓ' = 2×0.100 t = 0 ⇒ t = 0
Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s
⇒ vertex = (0,3.00)
7) The domain is given t ∈ [0,5.00]
8) You can also build a table with several points in the domain
t =0; Vₓ = 3.00 + 0.100 (0)² = 0
t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10
t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40
t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90
t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60
t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50
9) Range: Vₓ ∈ [ 3.00, 5.50]
10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.
