Answer:
R = 45.05 ft, θ = -29º
Explanation:
Let's analyze your problem, two cars collide that get stuck, then this collide with a third car and they get stuck and the whole collides with a wall stopping, this problem has several parts
We define a form system for all the cars, so that during the crashes the forces have been internal and the moment can be preserved
Let's slow down to the English system
vₐ₀ = 63 mi / h (5280 ft / 1 mile) (1 h / 3600 s) = 92.4 ft / s
v_{bo] = 76.8757 mi / h = 112.7 ft / s
v_{c0} = 45 mi / h = 66 ft / s
Let's start by looking for the velocity in the first collision, let's set a reference system with the x-axis in the eastern direction
X axis
initial instant. Before crash
p₀ = mₐ vₐ₀
final instant. Right after the crash
p_f = (mₐ + m_b) v₁ₓ
the moment will be preserved
p₀ = p_f
mₐ vₐ₀ = (mₐ + m_b) v₁ₓ
v₁ₓ =
let's look for speed
note that the unit of free is the weight (W = mg), but since it is in the numerator and denominator the value of gravity is eliminated, so we do not have to reduce this unit
v₁ₓ =
v₁ₓ = 42 ft / s
Y axis
initial
p₀ = -m_b v_{bo}
the negative sign is because it travels south (negative of the y-axis)
final
p_f = (mₐ + m_b) v_{1y}
conservation moment
p₀ = p_f
-m_b v_{bo} = (mₐ + m_b) v_{1y}
v_{1y} =
we look for the value
v_{1y} =
v_{1y} = -51.227 ft / s
Now let's solve the second clash
X axis
initial
p₀ = (mₐ + m_b) v₁ₓ
final
p_f = (mₐ + m_b + m_c) v₂ₓ
conservation moment
(mₐ + m_b) v₁ₓ = (mₐ + m_b + m_c) v₂ₓ
M = mₐ + m_b + m_c
v₂ₓ =
we calculate
M = 3000 + 3600 + 3000 = 9600
v₂ₓ =
v₂ₓ = 26.25 ft / s
Y axis
initial
p₀ = (mₐ + m_b) v_{1y} + m_c v_{co}
final
m_f = M v_{2y}
conservation moment
(mₐ + m_b) v_[1y} + m_c v_{co} = M v_{2y}
v_{2y} =
we calculate
v_{2y} =
v_{2y} = - 35.22 + 20.625
v_{2y} = - 14.594 ft / s
We already have the velocity of the set of vehicles, now we can use the kinematics relations to find the diastase at point D
For this last part we must make some assumptions
* we despise collision times
* we find the distance that the given time advances (t = 1.5 s) using the equations of uniform motion
X axis
x = v₂ₓ t
x = 26.25 1.5
x = 39.375 ft
Axis y
y = v_{2y} t
y = -14.594 1.5
y = -21.89 ft
We can give the answer in two ways
* the position is 39.375 ft to the East and 21.89 ft to the South
* In the form of module and angle
Let's use the Pythagorean theorem
R =
R =
R = 45.05 ft
let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (-21.89 / 39.375)
θ = -29º
this angle is measured clockwise from the x axis