Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How
1 answer:
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
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Answer:
The answer to the question
The steady state response is u₂(t) = -cos(3t + π/4)
of the form R·cos(ωt−δ) with R = , ω = 3 and δ = -π/4
Explanation:
To solve the question we note that the equation of motion is given by
m·u'' + γ·u' + k·u = F(t) where
m = mass = 2.00 kg
γ = Damping coefficient = 1
k = Spring constant = 3 N·m
F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)
Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)
The homogeneous equation 2·u'' + u' + 3·u is first solved as follows
2·u'' + u' + 3·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by =
This gives the general solution of the homogeneous equation as
u₁(t) =
For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)
Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have
3·B-15·A = 27
3·A +15·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)
u₂(t) = 1.5·(sin(3·t) - cos(3·t) = ·cos(3·t + π/4)
The general solution is then u(t) = u₁(t) + u₂(t)
however since u₁(t) = ⇒ 0 as t → ∞ the steady state response = u₂(t) =
·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where
R =
ω = 3 and
δ = -π/4