Cars a and b are racing each other along the same straight road in following manner: car a has a head start and is a distance da beyond the starting line at t=0. the starting line is at x=0. car a travels at a constant speed va. car b starts at the starting line but has a better engine than car a, and thus car b travels at a constant speed vb, which is greater than va.(a) how long after car b started the race will car b catch up with car a? (b) how far from car b's starting line will the cars be when car b passes car a?

tA=tB

tB=(sA-dA)/vA watch it below

Explanation:

the spatio for A is sA=vA*tA+dA

the spatio for B is sB=vB*tB

the meeting point will happen at: tA=tB

so, sA=vA*tB+dA

tB=sB/vB

yields:

sA=vA*(sB/vB)+dA

sA-dA=vA*(sB/vB)

sB=(sA-dA)*vB/vA

then

tB=((sA-dA)*vB/vA)/vB

tB=(sA-dA)/vA

answer; ///i do believe that the correct answer is ; ///(physical change);

a.metals can become magnetic if they have many spinning electrons that create magnetic fields.d.only some metals are attracted by magnets; these metals are called ferromagnetic.

d.wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.