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Metamorphose
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1. Using a reference frame with the origin at the take-off airport, the positive x-axis due East,
and the positive y-axis due North, the acceleration a of an airplane as a function of time
can be described as:
a = (αt)(i) + (βt4 - γt)(j)
with α, β, and γ positive and constant.
Assuming that the airplane takes off from the airport at time t = 0 with zero initial
velocity:
a) What are the units of α, β, and γ?
b) Find the time(s) when the airplane position is directly NE of the airport.
c) Find the trajectory of the plane, y(x).
Write your results in terms of α, β, and γ. Remember to check the dimensions/units for each
answer.
a = (αt)(i) + (βt4 - γt)(j)
v = ∫a(dt) → (0.5αt2)(i) + (1/5(βt5) - 0.5γt2)(j)
r = ∫v(dt) → (1/6(αt3)(i) + (1/30(βt6 - 1/6(γt3)(j)
[a.] For part A,
the units for α, β, and γ respectively:
ms-3, ms-6, and ms-3
[b.] For part B, to find the times at which the plane is directly NE would mean setting the rx and ry components equal to each other.
∴rx = ry
1/6(αt3 = 1/30(βt6 - 1/6(γt3
1/6(αt3 + 1/6(γt3 = 1/30(βt6
Multiplying the entire thing by 30 gives:
5αt3[/SUP + 5γt3 = βt6
t3(5α + 5γ) = βt6
(5α + 5γ) = βt3
t3 = (5α + 5γ)/β
∴t = [(5α + 5γ)/β]1/3
[c.] For part C, I know trajectory means projectile motion, but I am not sure how to do it.
Do we simply get the time and plug it into ry?Doing that gives 5α(α - γ)/6β as a final answer.
and the positive y-axis due North, the acceleration a of an airplane as a function of time
can be described as:
a = (αt)(i) + (βt4 - γt)(j)
with α, β, and γ positive and constant.
Assuming that the airplane takes off from the airport at time t = 0 with zero initial
velocity:
a) What are the units of α, β, and γ?
b) Find the time(s) when the airplane position is directly NE of the airport.
c) Find the trajectory of the plane, y(x).
Write your results in terms of α, β, and γ. Remember to check the dimensions/units for each
answer.
Homework Equations
a = (αt)(i) + (βt4 - γt)(j)
v = ∫a(dt) → (0.5αt2)(i) + (1/5(βt5) - 0.5γt2)(j)
r = ∫v(dt) → (1/6(αt3)(i) + (1/30(βt6 - 1/6(γt3)(j)
The Attempt at a Solution
[a.] For part A,
the units for α, β, and γ respectively:
ms-3, ms-6, and ms-3
[b.] For part B, to find the times at which the plane is directly NE would mean setting the rx and ry components equal to each other.
∴rx = ry
1/6(αt3 = 1/30(βt6 - 1/6(γt3
1/6(αt3 + 1/6(γt3 = 1/30(βt6
Multiplying the entire thing by 30 gives:
5αt3[/SUP + 5γt3 = βt6
t3(5α + 5γ) = βt6
(5α + 5γ) = βt3
t3 = (5α + 5γ)/β
∴t = [(5α + 5γ)/β]1/3
[c.] For part C, I know trajectory means projectile motion, but I am not sure how to do it.
Do we simply get the time and plug it into ry?Doing that gives 5α(α - γ)/6β as a final answer.
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