This is a practice problem that I'm pretty sure I got correct. I got that she rode 3388 meters on the trip. A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part,she rides for 21.6 minutes at an average speed of 9.33 m/s. During the second part, she rides for 28.0 minutes at an average speed of4.52 m/s. Finally, during the third part, she rides for 12.3 minutes at an average speed of 18.4 m/s. (a) How far has the bicyclist traveledduring the entire trip? (b) What is the average speed of the bicyclist for the trip?(a) Number i 3.39E3(b) Number 8.45UnitsUnitsm/sm

Given that:During first part of trip:Distance of first trip is given by:During second part…

Answered: This is a practice problem that I'm…

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This is a practice problem that I'm pretty sure I got correct. I got that she rode 3388 meters on the trip.

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Question

This is a practice problem that I'm pretty sure I got correct. I got that she rode 3388 meters on the trip.

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part,
she rides for 21.6 minutes at an average speed of 9.33 m/s. During the second part, she rides for 28.0 minutes at an average speed of
4.52 m/s. Finally, during the third part, she rides for 12.3 minutes at an average speed of 18.4 m/s. (a) How far has the bicyclist traveled
during the entire trip? (b) What is the average speed of the bicyclist for the trip?
(a) Number i 3.39E3
(b) Number 8.45
Units
Units
m/s
m

Expert Solution

Step 1: Given Information

Given that:

During first part of trip:

$t i m e space equals space 21.6 space m i n u t e s equals space 21.6 space X space 60 space equals space 1296 space s e c$

$A v e r a g e space s p e e d space equals space 9.33 space m divided by s$

Distance of first trip is given by:

$D subscript 1 equals s p e e d cross times t i m e D subscript 1 equals 9.33 cross times 1296 D subscript 1 equals 12091.68 space m$

During second part trip:

$t i m e space equals space 28 space m i n u t e s equals space 28 space X space 60 space equals space 1680 space s e c$

$A v e r a g e space s p e e d space equals space 4.52 space m divided by s$

Distance of second trip is given by:

$D subscript 2 equals s p e e d cross times t i m e D subscript 2 equals 4.52 cross times 1680 D subscript 2 equals 7593.6 space m$

During third part trip:

$t i m e space equals space 12.3 space m i n u t e s equals 12.3 space X space 60 space equals 738 space s e c$

$A v e r a g e space s p e e d space equals space 18.4 space m divided by s$

Distance of third trip is given by:

$D subscript 3 equals s p e e d cross times t i m e D subscript 3 equals 18.4 cross times 738 D subscript 3 equals 13579.2 space m$

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