Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
[tex]x=x_o+v_1t[/tex] (1)
xo = 10 km
v1 = 55km/h
car 2:
[tex]x'=v_2t[/tex] (2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
[tex]x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}[/tex]
[tex]t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s[/tex]
The position in which both cars coincides is:
[tex]x=(55km/h)(0.33h)=18.33km[/tex]