Question

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 27.8 minutes at an average speed of 6.13 m/s. During the second part, she rides for 44.7 minutes at an average speed of 3.16 m/s. Finally, during the third part, she rides for 8.19 minutes at an average speed of 19.7 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?

Answer 1

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

A bicyclist makes a trip that consists of three parts, each inthe same direction (due north) along a straight road. Duringthe first part, she rides for 22 minutes at an average speed of 7.2m/s. During the second part she rides for 36 minutes at an averagespeed of 5.1 m/s. Finally during the third part she rides for 8.0minutes at an average speed of 13 m/s.

a) how far has the bicyclist traveled during the entiretrip?

b) what is her average velocity for the trip?

Break it down into parts

22 minutes * 60s/ minute = 1320 seconds

36 minutes * 60 s/minute = 2160 seconds

8 minutes * 60 s/minute = 480 seconds

Total time taken:

1320 + 2160 + 480 =

3960 s

__________________________________

We know d=r*t

a.) Calculate the distance at each juncture

d = 7.2(1320)                  = 9504   m

d = 5.1(2160)                 = 11016 m

d = 13(480)                     = 6240   m

Add the distances to find the total distance traveled

D = 9504 + 11016 + 6240 = 26760 m

The total distance travelled is 27000m    ***Note: two sigfigs

b.) The average velocity is total distance (in this case it isalso the total displacement beacuse you are onlytraveling in one direction) divided by total time

V = D/t

V = 26760 / 3960

V= 6.757575

The average velocity was 6.8 m/s   ***Note:two sigfigs