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Arlington and Vernonville are 160 miles apart. A car leaves Arlington traveling towards​ Vernonville, and another car leaves Vernonville at the same​ time, traveling towards Arlington. The car leaving Arlington averages 10 miles per hour more than the​ other, and they meet after 1 hour and 36 minutes. What are the average speeds of the​ cars?1. The car leaving Arlington averages_______MPH2. The car leaving Vernonville averages_______MPH

User Poy Chang
by
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2 Answers

3 votes

Answer:Arlington moves at 55miles/hr

Vernon ville moves at 45 miles/hr

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

User Mzq
by
4.5k points
7 votes

Answer:

The car leaving Arlington averages 55 mph.

The car leaving Vernonville averages 45 mph.

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!

User Rito
by
5.5k points