Why does bromine add to a double bond instead of a triple bond if both are present?
I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater. 

A Kinetic Perspective

The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the pi-bond donates to the sigma* antibonding orbital to form a sigma C-Br bond. In doing so, both carbons in the alkene/alkyne will have their p-orbitals hybridized with the sp/sp2 orbital to form the sp2/sp3 orbital respectively. However, the sp2 bonding orbital will have more ring strain, as it naturally exists with bond angles of 120 degrees away from other sp2 orbitals, while the sp3 orbital is somewhat less strained with bond angles ~109.5 degrees apart. Thus the higher energy transition state will make the alkyne less reactive, and bromination occurs on the alkene first. 



In addition, (as was pointed out here : https://www.masterorganicchemistry.com/2013/05/29/alkyne-addition-pathways-the-3-membered-ring-pathway/) the p orbitals on Cl conjugate to the pi-bond on the alkene transition state, resulting in a 4-electron anti-aromatic system! This further increases the instability of the transition state and results in a slower reaction.

TL;DR: There's more ring strain on the alkyne's transition state because the carbons are sp2 hybridized, and its transition state involves the formation of an anti-aromatic ring, so the reaction occurs slower than bromination on the alkene.