Respuesta :
Given:
Total distance = 60miles
Ken speed 40 mph faster than Joe.
Joe take 2 hours longer than Ken.
Find-:
Joes speed
Sol:
Formula of speed is:
[tex]\text{ Speed = }\frac{\text{ Distance}}{\text{ Time}}[/tex]Let Joe's speed is "x"
Then Ken's speed is "x+40"
Let Ken take time is "t"
Then Joe takes time is "t+2"
For Joe's speed:
[tex]\begin{gathered} \text{ Speed = }x \\ \\ \text{ Time = }t+2 \end{gathered}[/tex]So, speed is:
[tex]\begin{gathered} \text{ Speed = }\frac{\text{ Distance}}{\text{ Time}} \\ \\ x=\frac{60}{t+2}.................(1) \\ \\ \end{gathered}[/tex]For Ken's speed:
[tex]\begin{gathered} \text{ Speed = }x+40 \\ \\ \text{ Time = }t \end{gathered}[/tex]So, speed is:
[tex]\begin{gathered} \text{ Speed =}\frac{\text{ Distance}}{\text{ Time}} \\ \\ x+40=\frac{60}{t} \\ \\ t=\frac{60}{x+40}..............................(2) \end{gathered}[/tex]From eq(2) put the value of "t" in eq(1) then:
[tex]\begin{gathered} x=\frac{60}{t+2} \\ \\ t+2=\frac{60}{x} \\ \\ t=\frac{60}{x}-2 \\ \\ \frac{60}{x+40}=\frac{60}{x}-2 \end{gathered}[/tex]Then, solve for "x"
[tex]\begin{gathered} \frac{60}{x+40}=\frac{60}{x}-2 \\ \\ \frac{60}{x+40}=\frac{60-2x}{x} \\ \\ 60x=(x+40)(60-2x) \\ \\ 60x=60x-2x^2+2400-80x \\ \\ 2x^2+80x-2400=0 \\ \\ x^2+40x-1200=0 \end{gathered}[/tex]Solve the quadratic equation then:
[tex]\begin{gathered} x^2+40x-1200=0 \\ \\ x^2+60x-20x-1200=0 \\ \\ x(x+60)-20(x+60)=0 \\ \\ (x+60)(x-20)=0 \\ \\ x=-60,20 \end{gathered}[/tex]Negative speed not possible so Joe speed is 20 mph
