Jacob drove from town a to town b at an average rate of x miles per hour, then returned along the same route at y miles per hour. if he then drove back to town b at z miles per hour along the same route, what was jacob's average rate of speed for the entire trip, in miles per hour?
Accepted Solution
A:
Average speed = (total distance traveled)/(total travel time) = (total distance)/(time of 1st journey + time of 2nd journey + time of 3rd journey)
Let d = the distance between Town A and Town B So, total distance traveled = 3d
Time = distance/speed time of 1st journey = d/x time of 2nd journey = d/y time of 3rd journey = d/z
Total time = d/x + d/y + dz To simplify, rewrite with common denominator: dyz/xyz + dxz/xyz + dxy/xyz So, total time = (dyz + dxz + dxy)/xyz
Average speed = (total distance)/(total time) = 3d/[(dyz + dxz + dxy)/xyz] = (3dxyz)/(dyz + dxz + dxy) Divide top and bottom by d to get: (3xyz)/(yz + xz + xy)