Question 1126541


Koech left home to a shopping center {{{12km}}} away, running at {{{8(km/h)}}}.

{{{d=12km}}}
{{{s=8(km/h)}}}
{{{12km=8(km/h)*t}}}..........eq.1

fifteen minutes later, mutual left the same home and cycled to the shopping center at {{{20(km/h)}}}

{{{15min=(1/4)h=0.25h }}}

{{{s=20( km/h)}}}

time is {{{t-0.25h}}}

{{{d}}} is same: {{{d=12km}}}

{{{12km=20( km/h)(t-0.25h)}}}....eq.2

they will meet when

{{{8(km/h)*t=20( km/h)(t-0.25h)}}} solve for {{{t}}}

{{{8cross((km/h))*t=20cross(( km/h))(t-0.25h)}}}

{{{8t=20(t-0.25h)}}}

{{{8t=20t-20*0.25h}}}

{{{8t=20t-5h}}}

{{{20t-8t=5h}}}

{{{12t=5h}}}

{{{t=(5/12)h}}}

{{{t}}}≈{{{0.416667h}}}

ind distance:

{{{d=8(km/h)*t}}}

{{{d=8(km/h)*0.416667h}}}

{{{d=3.333333km}}}-> Koech

{{{d=20( km/h)(0.416667h-0.25h)}}}

{{{d=20( km/h)(0.166666667h)}}}

{{{d=3.3333333km}}}->the distance to the shopping center at which Mutua caught up with Koech