Cruiser A was traveling east at 63 mi/h on a police emergency call when it was hit at an intersection by car B, which was traveling south at high speed. After sliding together on the wet pavement, the two cars hit cruiser C, which was traveling north at 45 mi/h and was 63 ft from the intersection at the time of the first collision. Stuck together, the three cars hit a wall and came to a stop at D. Knowing that car B weighs 3600 lb and that each of the cruisers weighs 3000 lb, solve the problem by neglecting the forces exerted on the cars by the wet pavement and treating the cars as point masses. Knowing that the speed of car B was 76.8757 mi/h and that the time elapsed from the first collision to the stop at D was 1.5 s, determine the coordinates of D.

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 = [tex]\frac{m_a}{m_a+m_b} \ v_{ao}[/tex]

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 = [tex]\frac{3000}{3000+3600} \ 92.4[/tex]

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} = [tex]- \frac{m_a}{m_a+m_b} \ v_{1y}[/tex]

we look for the value

v_{1y} = [tex]- \frac{3000}{3000+3600} \ 112.7[/tex]

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 = [tex]\frac{m_a+m_b}{M} \ v_{1x}[/tex]

we calculate

M = 3000 + 3600 + 3000 = 9600

v₂ₓ = [tex]\frac{ 3000+3600}{9600} \ 42[/tex]

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} = [tex]\frac{m_a+m_b}{M } \ v_{1y} + \frac{m_c}{M} \ v_{co}[/tex]

we calculate

v_{2y} = [tex]- \frac{3000+3600}{9600} \ 51.227 + \frac{3000}{9600} \ 66[/tex]

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 = [tex]\sqrt{x^2+ y^2}[/tex]

R = [tex]\sqrt{39.375^2 + 21.89^2}[/tex]

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