Q:

Two trains left the cities A and B at the same time and headed towards each other. The cities are s miles apart. The first train was traveling at the speed of v1 mph. The second train was traveling at the speed of v2 mph. After t hours the two trains met each other. Find the formula for t in terms of s,v1 and v2. Then find the value of t if:as=250, v1 = 60 , v2 = 40The formula for t is: (A) s · (v1 +v2) (B) s v1 +v2 (C) s v2 −v1 (D) v1 +v2 s (E) (v2 −v1) ·s

Accepted Solution

A:
The formula of t in terms of s , [tex]v_{1}[/tex] , [tex]v_{2}[/tex] is [tex]t=\frac{s}{(v_{1}+v_{2})}[/tex]The value of t  when s = 310 , [tex]v_{1}[/tex] = 75 , [tex]v_{2}[/tex] = 80 is 2 hoursStep-by-step explanation:Two trains left the cities A and B at the same time and headed towardseach other The cities are s miles apartThe first train was traveling at the speed of [tex]v_{1}[/tex] mphThe second train was traveling at the speed of [tex]v_{1}[/tex] mphAfter t hours the two trains met each other∵ The cities are s miles apart∵ After t hours the two trains met each other∴ s = [tex]s_{1}[/tex] + [tex]s_{2}[/tex], where [tex]s_{1}[/tex] is the    distance that the 1st train moved in t hours and [tex]s_{2}[/tex]     is the distance that the 2nd train moved in t hours∵ s = v t∴ [tex]s_{1}[/tex] = [tex]v_{1}[/tex] t∴ [tex]s_{2}[/tex] = [tex]v_{2}[/tex] t- Substitute them in the equation of s∴ s = [tex]v_{1}[/tex] t + [tex]v_{2}[/tex] t- Take t as a common factor in the right hand side∴ s = t ( [tex]v_{1}[/tex] + [tex]v_{2}[/tex] )- Divide both sides by ( [tex]v_{1}[/tex] + [tex]v_{2}[/tex] )∴ [tex]t=\frac{s}{(v_{1}+v_{2})}[/tex]The formula of t in terms of s , [tex]v_{1}[/tex] , [tex]v_{2}[/tex] is [tex]t=\frac{s}{(v_{1}+v_{2})}[/tex]∵ s = 310 m/h∵ [tex]v_{1}[/tex] = 75 m/h∵ [tex]v_{2}[/tex] = 80 m/h- Substitute these values in the formula of t above∴ [tex]t=\frac{310}{(75+80)}=\frac{310}{155}[/tex]∴ t = 2 hoursThe value of t  when s = 310 , [tex]v_{1}[/tex] = 75 , [tex]v_{2}[/tex] = 80 is 2 hoursLearn more:You can learn more about distance, speed and time in brainly.com/question/1748290#LearnwithBrainly