Question #83f7e

A #stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxx))(color(red)(rarr)# X#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(blue)(larr)# B
#" "d_A" "d_B#

We use the fact that both runs share the same time t.

We have 3 unknowns and we can set up 3 simultaneous equations to solve them.

We know that average velocity = distance travelled / by the time taken.

#d_A# = distance travelled by A

#d_B# = distance travelled by B

#t# is the time taken

#stackrel(rarr)(V_A)=d_A/t#

#:.7=d_A/t" "color(red)((1))#

#stackrel(rarr)(V_B)=d_B/t#

#:.6=d_B/t" "color(red)((2))#

and we know that:

#d_A+d_B=220" "color(red)((3))#

#:.d_A=(220-d_B)#

Substituting this into #color(red)((1))rArr#

#7=((220-d_B))/t" "color(red)((4))#

From #color(red)((2))# we get #d_B=6t#

Substituting this into #color(red)((4))rArr#

#7=((220-6t))/t#

#:.7t=220-6t#

#t=220/13=16.92color(white)(x)s#

This is the time taken for the runners to meet.

Rearranging #color(red)((1))rArr#

#d_A=7t#

#d_A=7xx16.92=118.46color(white)(x)m#

This is the distance from point A where they meet.

Check the distance that B has run:

#d_B=220-d_A=220-118.46=101.53color(white)(x)m#

As you would expect runner A covers more distance than runner B as he/she is running faster.