Respuesta :
A) Car A is initially ahead
B) The two cars are at the same point at the times: t = 0, t = 2.27 s and
t = 5.73 s
C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s
D) The two cars have same acceleration at t = 2.67 s
Explanation:
A)
The position of the two cars at time t is given by the following functions:
[tex]x_A(t) = \alpha t + \beta t^2[/tex]
with
[tex]\alpha = 2.60 m/s\\\beta = 1.20 m/s^2[/tex]
Substituting,
[tex]x_A(t)=2.60t+1.20 t^2[/tex]
And
[tex]x_B(t)=\gamma t^2 - \delta t^3[/tex]
with
[tex]\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3[/tex]
Substituting,
[tex]x_B(t)=2.80t^2-0.20t^3[/tex]
Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:
[tex]x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m[/tex]
[tex]x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m[/tex]
So, car A is initially ahead.
B)
The two cars are at the same point when their position is the same. Therefore, when
[tex]x_A(t)=x_B(t)[/tex]
which means when
[tex]2.60t+1.20t^2 = 2.80t^2-0.20t^3[/tex]
Re-arranging the equation, we find
[tex]0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0[/tex]
One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation
[tex]0.20t^2-1.6t+2.60=0[/tex]
which has two solutions:
t = 2.27 s
t = 5.73 s
So, these are the times at which the cars are at the same point.
C)
The distance between the two cars A and B is not changing when the velocities of the two cars is the same.
The velocity of car A is given by the derivative of the position of car A:
[tex]v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t[/tex]
The velocity of car B is given by the derivative of the position of car B:
[tex]v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2[/tex]
Therefore, the distance between the two cars is not changing when the two velocities are equal:
[tex]v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0[/tex]
This is another second-order equation, which has two solutions:
t = 1.00 s
t = 4.33 s
D)
The acceleration of each car is given by the derivative of the velocity of the car A.
The acceleration of car A is:
[tex]a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40[/tex]
While the acceleration of car B is:
[tex]a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t[/tex]
So, the two cars have same acceleration when
[tex]a_A(t)=a_B(t)[/tex]
And solving the equation, we find:
[tex]2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s[/tex]
So, the two cars have same acceleration at t = 2.67 s.
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The given functions for the distances of car A and car B from
the starting point, are functions with graphs that intersect.
Correct responses:
Part A: Car A
Part B: t = 0, t = [tex]\underline{(4 - \sqrt{3} )}[/tex], t = [tex]\underline{(4 + \sqrt{3}) }[/tex]
Part C: t = 13 s, and [tex]\underline{t = \dfrac{1}{3} \, s }[/tex]
Part D: [tex]\underline{2. \overline{6} \ s}[/tex]
How can the distance, time, and acceleration be found from functions
Distance of car A from the starting point, [tex]x_A(t)[/tex] = α·t + β·t²
α = 2.60 m/s
β = 1.20 m/s²
Distance of car B from the start, [tex]x_B(t)[/tex] = γ·t² - δ·t³
γ = 2.80 m/s²
δ = 0.20 m/s³
Part A
The velocity of each car at the start can give the car which is
ahead from the start.
[tex]Velocity \ of \ car \ A = x_A'(t) = \mathbf{\alpha + 2 \cdot \beta \cdot t}[/tex]
Therefore, at the start, t = 0, which gives;
Velocity of car A = 2.60 + 2 × 1.20 × 0 = 2.60
Velocity of car A at the start is 2.60 m/s
[tex]Velocity \ of \ car \ B = \mathbf{x_B'(t)} = 2 \cdot \gamma \cdot t - 3 \cdot \delta \cdot t^2[/tex]
At the start, [tex]x_B'(t)[/tex] = 2 × 2.80 × 0 - 3 × 0.20 × 0 = 0
Velocity of car B at the start is 0
Therefore, car A is faster than car B at the start
Therefore;
- Car A is ahead of car B just after they leave the starting point.
Part B
When the cars are at the same point, we have;
[tex]\mathbf{x_A(t)}[/tex] = [tex]x_B(t)[/tex]
Which gives;
α·t + β·t² = γ·t² - δ·t³
From which we have;
2.60·t + 1.20·t² = 2.80·t² - 0.20·t³
Which gives;
0.20·t³ - 2.80·t² + 1.20·t² + 2.60·t = 0
0.20·t³ - 1.60·t² + 2.60·t = 0
Which gives;
t·(0.20·t² - 1.60·t + 2.60) = 0
Therefore, t = 0, is a solution
0.20·t² - 1.60·t + 2.60 = 0
Dividing by 0.20 gives;
0.20·(t² - 8·t + 13) = 0
t² - 8·t + 13 = 0
Which gives;
[tex]t = \dfrac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times 13} }{2 \times 1} = 4 \pm \dfrac{\sqrt{4 \times 3} }{2} = 4 \pm \sqrt{3}[/tex]
The times at which the cars are at the same point are;
- [tex]\underline{t = 0 \ s, \ t = (4 - \sqrt{3}) \ s, \ t = (4 + \sqrt{3}) \ s }[/tex]
Part C
The distance of car A from B is; d = [tex]\mathbf{x_B(t)}[/tex] - [tex]\mathbf{x_A(t)}[/tex]
Which gives;
d = (γ·t² - δ·t³) - (α·t + β·t²)
When the distance is neither increasing or decreasing, is at the extremum point, which gives;
d = (2.80·t² - 0.20·t³) - (2.60·t + 1.20·t²)
At the extremum point, d' = 0
Therefore;
d' = (5.60·t - 0.60·t²) - (2.60 - 2.40·t) = 0
0.60·t² - (5.60 + 2.40)·t + 2.60 = 0
0.60·t² - 8·t + 2.60 = 0
Which gives;
[tex]t = \mathbf{\dfrac{8 \pm \sqrt{(-8)^2 - 4 \times 0.60 \times 2.60} }{2 \times 0.6}} = \dfrac{40}{6} \pm \dfrac{\sqrt{57.76 } }{1.2} =\dfrac{20}{3} \pm \dfrac{19}{3}[/tex]
Which gives;
[tex]t = \dfrac{39}{3} = 13[/tex], [tex]t = \dfrac{20 - 19}{3} = \dfrac{1}{3}[/tex]
The times at which the distance between A and B is neither increasing nor decreasing are;
- t = 13 s, and [tex]\underline{t = \dfrac{1}{3} \, s}[/tex]
Part D
The acceleration are;
[tex]x''_A(t)[/tex] = 2·β
[tex]x''_B(t)[/tex] = 2·γ - 6·δ·t
When the acceleration are the same, we have;
[tex]x''_A(t)[/tex] = [tex]\mathbf{x''_B(t)}[/tex]
Which gives;
2·β = 2·γ - 6·δ·t
2 × 1.2 = 2 × 2.80 - 6 × 0.20 × t
2.4 = 5.60 - 1.20·t
1.20·t = 5.60 - 2.40 = 3.2
[tex]t = \dfrac{3.20}{1.20} = \dfrac{8}{3} = 2\frac{2}{3} = 2.\overline 6[/tex]
- The time at which A and B have the same acceleration is [tex]\underline{t = 2. \overline 6} \ s[/tex]
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