Answer:
see below
Step-by-step explanation:
Let t = time (in hours)
a) i) First 105 km of journey
[tex]\mathsf{speed=\dfrac{distance}{time}}[/tex]
[tex]\implies x=\dfrac{105}{t_1}[/tex]
[tex]\implies t_1=\dfrac{105}{x}[/tex]
ii) Rest of journey
distance = 190 - 105 = 85 km
speed = [tex]x+12[/tex] km/h
[tex]\mathsf{speed=\dfrac{distance}{time}}[/tex]
[tex]\implies x+12=\dfrac{85}{t_2}[/tex]
[tex]\implies t_2=\dfrac{85}{x+12}[/tex]
b) 3 hours 15 minutes = 3.25 hours
[tex]\implies[/tex] total time = [tex]t_1+t_2=3.25[/tex]
[tex]\implies \dfrac{105}{x}+\dfrac{85}{x+12} =3.25[/tex]
[tex]\implies \dfrac{105(x+12)+85x}{x(x+12)} =3.25[/tex]
[tex]\implies 105(x+12)+85x =3.25x(x+12)[/tex]
[tex]\implies 105x+1260+85x =3.25x^2 +39x[/tex]
[tex]\implies 3.25x^2-151x-1260=0[/tex]
multiply by 4:
[tex]\implies 13x^2-604x-5040=0[/tex]
c)
[tex]\textsf{quadratic formula} \ x=\dfrac{-b \pm\sqrt{b^2-4ac} }{2a}}[/tex]
[tex]\implies x=\dfrac{604 \pm\sqrt{(-604)^2-4(13)(-5040)} }{2(13)}}[/tex]
[tex]\implies x=\dfrac{604 \pm\sqrt{626896} }{26}}[/tex]
[tex]\implies x = 53.68, x=-7.22[/tex]
(to 2 decimal places)
d) time ≥ 0, therefore, x = 53.68 only
Substitute x = 53.68 into the expression for [tex]t_1[/tex] and solve for [tex]x[/tex]:
[tex]\implies t_1=\dfrac{105}{53.68}=1.955912... \textsf{hours}=\textsf{1 hr 57 min}[/tex]
