After #t# hours, Sarah would be
#t " hr" xx 10 " km/h"#
# = 10t " km"#
North of the starting point;
and
(provided #t>=1#) Becca would be
#t " hr" xx 15 " km/hr"#
#= 15(t-1) " km"#
East of the starting point.
Using the Pythagorean Theorem, the distance between them would be:
#color(white)("XXX")D= sqrt((10t)^2+(15(t-1))^2)#
#color(white)("XXXX")= sqrt(100t^2+225(t^2-2t+1))#
#color(white)("XXXX")=sqrt(325t^2-450t+225)#
#color(white)("XXXX")=5(13t^2-18t+13)^(1/2)#
The rate of change of the distance between them would be:
#color(white)("XXX")(dD)/(dt) = 5(dD)/(dt)#
#color(white)("XXXX")= 5*1/2(13t^2-18t+13)^(-1/2)*(26t-18)#
#color(white)("XXXX")= (65t-45)/sqrt(13t^t-18t+13)#
At 9 am (time #t=2# based on starting time of 7 am)
The rate of change would be:
#color(white)("XXX")(65(2)-18)/sqrt(13*(2)^2-18t+13)#
#color(white)("XXX")~~20.8#
