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A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 10.0 min at 75.0 km/h, 6.0 min at 95.0 km/h, and 45.0 min at 40.0 km/h and spends 40.0 min eating lunch and buying gas.(a) Determine the average speed for the trip.

(b) Determine the distance between the initial and final cities along the route.

User SuperUser
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1 Answer

10 votes

Answer:

a) v = 0.515 km / min , b) x_total = 52 km

Step-by-step explanation:

The measured speed is defined by the distance traveled between the time

v =
(\Delta x)/(\Delta t)

In this case they give us the speed in several time intervals

let's find the distance traveled in each interval

a) Goes at 75 km/h for 10 min

v =
(x)/(t)x / t

x₁ = v t

let's reduce speed to km / min

v₁ = 75 km / h (1h / 60 min) = 1.25 km / min

the distance traveled in this time is

x₁ = 1.25 10

x₁ = 12.5 km

b) goes to v = 95 km / h for 6 min

v = 95 km / h (1h 60 min) = 1.5833 km / min

the distance traveled is

x₂ = v₂2 t

x₂ = 1.58333 6

x₂ = 9.5 km

c) goes to v = 40 km / h for 45.0 min

v₃ = 40 km / h (1 h / 60min) = 0.6667 km / min

x₃ = 0.6667 45

x₃ = 30 km

d) t = 40 min, stopped

x₄ = 0

A) let's calculate the average speed of the trip

v =
(x_(1)+x_(2)+x_(3)+x_(4) )/(t_(1)+t_(2)+t_(3)+t_(4) )

v = (12.5 +9.5 +30 +0) / (10 +6 +45 +40)

v = 52/101

v = 0.515 km / min

B) the distance between the two cities is

x_total = x₁ + x₂ + x₃

x_total = 12.5 +9.5 + 30

x_total = 52 km

User James Brightman
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