- #1
Metamorphose
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- 0
1. Two cars, an Acura and a Bentley, are initially at rest a distance L from each other. At time t = 0 both cars starts accelerating toward each other, with constant acceleration Aa amd Ab respectively, until they crash into each other.
a) At what time does the crash happen?
b) Find the distance traveled by the Bentley before the crash.
c) What is the relative speed of the Bentley, with respect to the Acura, just before the
crash?Write your results in terms of L, Aa, and Ab. Remember to check the dimensions/units for
each answer.
2.
EQUATIONS: (WE CANNOT USE MEMORIZED EQUATIONS, WE MUST DERIVE THEM GIVEN THE INFORMATION.
Va = ∫Aa(dt) --> Aa(t) + Voa
Xa = ∫Va(dt) --> 0.5Aa(t)^2 + Voa(t) + Xoa
Vb = ∫Ab(dt) --> Ab(t) + Vob
Xb = ∫Vb(dt) --> 0.5Ab(t)^2 + Vob(t) + Xob
[a] The cars will crash when their positions are the same ---> Xa = Xb.
0.5Aa(t)^2 + Voa(t) + Xoa = 0.5Ab(t)^2 + Vob(t) + Xob
You can take out both Voa and Vob because the problem states that they are initially at rest. Let Xoa be at the origin, making it 0 and Xob = L because that is the distance between it and the Acura.∴ 0.5Aa(t)^2 = 0.5Ab(t)^2 + L
Aa(t)^2 - Ab(t)^2 = 2L ---> t^2 (Aa - Ab) = 2L
t = [(2L/(Aa - Ab))]^0.5. Discard the negative solution for time.[b.] Find the distance traveled by the Bentley before the crash.
Xb = 0.5Ab(t)^2 + Vob(t) + Xob
Once again Vob(t) = 0,
Xb = 0.5Ab([(2L/(Aa - Ab))]^0.5)^2 + L
Xb = 0.5Ab(2L/(Aa-Ab)) + L
Xb = [Ab(L)/(Aa-Ab)] + L
Is this correct?
Any help for this and part C will be greatly appreciated :)
a) At what time does the crash happen?
b) Find the distance traveled by the Bentley before the crash.
c) What is the relative speed of the Bentley, with respect to the Acura, just before the
crash?Write your results in terms of L, Aa, and Ab. Remember to check the dimensions/units for
each answer.
2.
EQUATIONS: (WE CANNOT USE MEMORIZED EQUATIONS, WE MUST DERIVE THEM GIVEN THE INFORMATION.
Va = ∫Aa(dt) --> Aa(t) + Voa
Xa = ∫Va(dt) --> 0.5Aa(t)^2 + Voa(t) + Xoa
Vb = ∫Ab(dt) --> Ab(t) + Vob
Xb = ∫Vb(dt) --> 0.5Ab(t)^2 + Vob(t) + Xob
The Attempt at a Solution
[a] The cars will crash when their positions are the same ---> Xa = Xb.
0.5Aa(t)^2 + Voa(t) + Xoa = 0.5Ab(t)^2 + Vob(t) + Xob
You can take out both Voa and Vob because the problem states that they are initially at rest. Let Xoa be at the origin, making it 0 and Xob = L because that is the distance between it and the Acura.∴ 0.5Aa(t)^2 = 0.5Ab(t)^2 + L
Aa(t)^2 - Ab(t)^2 = 2L ---> t^2 (Aa - Ab) = 2L
t = [(2L/(Aa - Ab))]^0.5. Discard the negative solution for time.[b.] Find the distance traveled by the Bentley before the crash.
Xb = 0.5Ab(t)^2 + Vob(t) + Xob
Once again Vob(t) = 0,
Xb = 0.5Ab([(2L/(Aa - Ab))]^0.5)^2 + L
Xb = 0.5Ab(2L/(Aa-Ab)) + L
Xb = [Ab(L)/(Aa-Ab)] + L
Is this correct?
Any help for this and part C will be greatly appreciated :)