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A person travels by car from one city to another with different constant speeds between pairs of cities. She drive 60.0 min at 100.0km/h, 5.0 min at 90.0 km/h, and 40.0 min at 60.0km/h and soends 30.0 min eating lunch and buying gas. (A) determine the average speed for the trip. (B) determine the distance between the intial and final cities alkng the route

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CintiaSalazar

Answer:

The average speed for the trip is 65.56 [tex]\frac{km}{h}[/tex] and the distance between the intial and final cities along the route is 147.5 km

Explanation:

You know:

  • t1= 60.0 min= 1 h
  • v1= 100.0km/h
  • t2= 5.0 min= 1/12 h
  • v2= 90.0 km/h
  • t3= 40.0 min= 2/3 h
  • v3= 60.0km/h
  • t4=30.0 min= 0.5 h= 1/2 h
  • v4= 0 because she stops to eat lunch and buy gas.

Speed is a quantity that reflects the space traveled by a body in a unit of time. That is, the average velocity relates the change in position with the time taken to effect said change.

[tex]speed=\frac{change in position}{time}[/tex]

To calculate the average speed of the trip you must calculate the total distance traveled and the total time used.

The total time used for the trip is:

t= t1 + t2 + t3 + t4= 60 min + 5 min + 40 min + 30 min= 135 min= 2.25 h

The distance traveled in each stage of the trip is calculated knowing the speed in that stage and the time used by the expression:

distance= speed*time

Then:

  • distance1= v1*t1 = 100 km/h * 1 h= 100 km
  • distance2= v2*t2 = 90 km/h * 1/12 h= 7.5 km
  • distance3= v3*t3 = 60 km/h * 2/3 h= 40 km
  • distance4= v4*t4 = 0 km/h * 1/2 h= 0 km

So, the total distance traveled is calculated as:

d= 100 km + 7.5 km + 40 km + 0 km= 147.5 km

Then, the average speed for the trip is:

[tex]average speed=\frac{total distance}{total time}[/tex]

[tex]average speed=\frac{147.5 km}{2.25 h}[/tex]

average speed= 65.56 [tex]\frac{km}{h}[/tex]

The average speed for the trip is 65.56 [tex]\frac{km}{h}[/tex] and the distance between the intial and final cities along the route is 147.5 km

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