Answer:
The car leaving Arlington averages 55 mph.
The car leaving Vernonville averages 45 mph.
Step-by-step explanation:
Hi there!
The traveled distance can be calculated as follows:
d = v · t
Where
d = traveled distance at time t.
v = speed.
t = time.
The cars meet after 1 h and 36 min, let´s convert it to hours only:
36 min · (1 h/60 min) = 0.6 h
Then, the cars meet at t = 1.6 h
We have the following equations:
d₁ + d₂ = 160 miles
d₁ = v₁ · 1.6 h
d₂ = v₂ · 1.6 h (v₂ = v₁ - 10 mph)
d₂ = (v₁ - 10 mph) · 1.6 h
Where:
d₁ = distance traveled by the car leaving Arlington
d₂ = distance traveled by the car leaving Vernonville.
v₁ = speed of the car leaving Arlington.
v₂ = speed of the car leaving Vernonville.
Then, let´s solve this system of equations:
d₁ + d₂ = 160 miles
v₁ · 1.6 h + (v₁ - 10 mph) · 1.6 h = 160 mi
divide both sides of the equation by 1.6 h:
v₁ + v₁ - 10 mph = 100 mph
add 10 mph to both sides of the equation:
2v₁ = 110 mph
divide by 2 the both sides of the equation:
v₁ = 55 mph
v₂ = v₁ - 10 mph = 55 mph - 10 mph = 45 mph
The car leaving Arlington averages 55 mph.
The car leaving Vernonville averages 45 mph.
Have a nice day!
Answer:Arlington moves at 55miles/hr
Vernon ville moves at 45 miles/hr
Step-by-step explanation:
Let a be the speed of the car leaving Arlington
Let v be the speed of the car leaving
Vernon ville
a = v+10
Distance travelled by the car leaving Arlington after 1hr 36minutes(1.6hrs)
= speed × time = 1.6a
Distance travelled by the car leaving Vernon ville after 1hr 36minutes(1.6hrs)
= speed × time = 1.6v
They meet after travelling this distance each thus covering 160 miles
1.6a + 1.6v = 160 --------1
a = v +10 (a is faster with 10miles/hr)
Put a in eqn 1
1.6(v+10) + 1.6v = 160
1.6v + 16 + 1.6v = 160
3.2v = 160-16=144
v =45 miles/hr
a =45+10= 55miles/hr
