Jaylen and ken started jogging from the same place in the opposite directions along a straight path. jaylen’s speed was 30 m/min faster than ken’s speed. both of them did not change their speeds throughout the jog. at the end of the jog, they were 13.6 km apart. jaylen jogged 2.4 km more than ken. what was ken’s average speed?

averagespeedofKenis 70 m/min. Using therelationbetween thespeedanddistance, the requiredspeedforKenis calculated.## What is the relation between speed and distance?

relationshipbetweenspeedS anddistanceD with aperiodT is formed asS = D/Tm/sec(basic unit)## Calculation:

JaylenandKenstarted jogging from thesameplace inoppositedirectionsalong a straight path.sumofdistancestraveled by both of them inoppositedirections = 13.6 km (since it is given that at the end of their jogging, they are 13.6 km apart)distancetraveled byJaylen= x kmdistancetraveled byKen= x – 2.4 km (sinceJaylenjogged 2.4 kmmorethanKen)Step 1: Finding the distance traveled by them:distancetravelled byJaylen=8kmand byKen= 8 – 2.4 =5.6kmStep 2: Finding the average speed traveled by Ken:Jaylen’sspeedwas 30 m/minfasterthanKen’sspeed.averagespeedofKenS = D/tS = 70 m/min.averagespeedofJaylenwill be 70 m/min + 30 m/min = 100 m/min.averagespeedofKenis 70 m/min.relationbetweenspeed,distance, andtimehere: