Question

Two runners start from the same point at the same time. They will be 8 km apart at the end of two hours if running in the same direction and they will be 26 km apart at the end of the one hour if running in opposite direction. find their speed?

Answer 1

Answer:

Let x = speed of the faster runner

Let y = speed of the slower

:

Write a distance equation for each scenario

:

2(x - y) = 4; running the same direction

simplify, divide by 2

x - y = 2

and

1(x + y) = 16; running opposite directions

:

Add the two equations

x - y = 2

x + y = 16

-------------addition eliminates y, find x

2x = 18

x = 9 mph is the faster runner

then

9 + y = 16

y = 16 - 9

y = 7 mph is the slower runner

:

:

Check solution in the two equations

2(9 - 7) = 4

and

1(9 + 6) = 16

Answer 2

Let the speeds of the runners be $\sf{v_1}$ and $\sf{v_2}$ respectively in $\sf{km\,h^{-1}}$ where $\sf{v_1>v_2.}$ Assume they move with uniform speed.

When they move in same direction the relative velocity between them is $\sf{v_1-v_2.}$

They will be 8 km apart from each other after 2 hours. So by second equation of motion,

$\sf{\longrightarrow 2(v_1-v_2)=8}$

$\sf{\longrightarrow v_1-v_2=4\ km\,h^{-1}\quad\quad\dots(1)}$

When they move in opposite directions the relative velocity between them is $\sf{v_1+v_2.}$

They will be 26 km apart from each other after 1 hour. So by second equation of motion,

$\sf{\longrightarrow 1(v_1+v_2)=26}$

$\sf{\longrightarrow v_1+v_2=26\ km\,h^{-1}\quad\quad\dots(2)}$

Solving (1) and (2) we get their speeds.

$\sf{\longrightarrow \underline{\underline{v_1=15\ km\,h^{-1}}}}$

$\sf{\longrightarrow \underline{\underline{v_2=11\ km\,h^{-1}}}}$