A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 45.0 min at 70.0 km/h, 9.0 min at 55.0 km/h, and 40.0 min at 60.0 km/h and spends 15.0 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.

Question
Asked Nov 21, 2019
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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 45.0 min at 70.0 km/h, 9.0 min at 55.0 km/h, and 40.0 min at 60.0 km/h and spends 15.0 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.

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Expert Answer

Step 1

Let the speed of the car in the first three interval is denoted by v1, v2 and v3 and for the corresponding time t1, t2 and t3.

Let tb be the time spent in eating and buying gas.

70 km/h, , = 45 min
v255 km/h, = 9 min
v 60 km/h,t, = 40 min
,=15 min
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70 km/h, , = 45 min v255 km/h, = 9 min v 60 km/h,t, = 40 min ,=15 min

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Step 2

(b) Use the speed-distance formula and plug the suitable values to determine the total distance(Δx) between final and the initial cities.

Ar = vh2+v,!3
|(70 km/h)(45 min)+(55 km/h)(9 min)-
(60 km/h)(40 min)
1h
60 min
= 100.75 km
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Ar = vh2+v,!3 |(70 km/h)(45 min)+(55 km/h)(9 min)- (60 km/h)(40 min) 1h 60 min = 100.75 km

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Step 3

Find the total time(Δt) ...

At 12 1
1 h
=[(45 min)+(9 min)+(40 min)+(15 min)
60 min
= 1.82 h
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At 12 1 1 h =[(45 min)+(9 min)+(40 min)+(15 min) 60 min = 1.82 h

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