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A cyclist completes a 100km race in 2:25:30. The race courseproceeded 50km north

ID: 1678815 • Letter: A

Question

A cyclist completes a 100km race in 2:25:30. The race courseproceeded 50km north, then 25km east, then 25km south. Her splittimes were 1:06:05 at the 50 km point and 1:49:19 at the 75kmpoint. a.) What is her displacement from start to finish (bothmagnitude and direction)? draw this to the diagram. b.)What is her average velocity from start to finish (both inmagnitude and direction)? c.) At the start fo the race the cyclist accelerates from 0 to15 m/s (north) in 7sec. What is her average acceleration (both inmagnitude and direction)? A cyclist completes a 100km race in 2:25:30. The race courseproceeded 50km north, then 25km east, then 25km south. Her splittimes were 1:06:05 at the 50 km point and 1:49:19 at the 75kmpoint. a.) What is her displacement from start to finish (bothmagnitude and direction)? draw this to the diagram. b.)What is her average velocity from start to finish (both inmagnitude and direction)? c.) At the start fo the race the cyclist accelerates from 0 to15 m/s (north) in 7sec. What is her average acceleration (both inmagnitude and direction)?

Explanation / Answer

Let the unit vector in east direction be i and in northdirection be j then displacement S = 50 km j + 25 km i +25km * ( -j)                                = 25 km j + 25 km i total displacment   S = [ 25^ 2 + 25 ^2]                               = 35.3553 km (b). Average velcoity V = S / total time                                      = 35.3553 km / ( 2h 2min 30 s)                                      = 35355.33 m / ( 2*3600s+120 s + 30 s)                                      = 35355.33 m / 7350 s                                      = 4.81 m / s (c).average accleration a = ( 15 - 0 ) / 7                                       =2.1428 m / s^ 2                                      
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