A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.

Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/sFind: AccelerationAnalysis:Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)50.41 m/s²=(70.8 m)aa=0.712 m/s²

Explanation:GiVeNA bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m. Determine the acceleration of the bikeT0 FInDAccelerationSoLuTi0NLet acceleration of the bike is a m/sec^2 bike started from rest so its initial speed will zero i.e.u=0It reaches at a speed of 7.10 m/s i.e finalspeed v= 7.10 m/sIn reaching speed 7.10 m/s ,it covered distance 35.4 m. Apply equation of motion which relatesu ,v,a and sV^2=u^2+2 as[tex]{ \rm \sf \pink{(7.10)^2=(0)^2+2 a (35.4)}}[/tex][tex]{ \rm \sf \green{50.41-70.8a}}[/tex][tex]{ \rm \sf \pink{a=0.712ms^(-2)}}[/tex]FiNaL AnsWeRSo acceleration of the bike which started from rest, reaches at speed of 7.10 m/s by travelling distance 35.4 m will be 0.712 m/ sec^2.