Question

Question #83f7e

Answer 1

They will meet at a point X 118 m from where A started after 17 s.

Explanation:

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.