Question

Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a
speed of 3.0 m/s and Betty has a speed of 2.0 m/s. Their dog, Spot, starts by Arthur's side at
the same time and runs back and forth between them at 5.0 m/s. By the time Arthur and
Betty meet, what distance has Spot run?

Answer 1

First find the time it took for Arthur and Betty to meet.


Use: \(x_f=x_i+v_0\Delta t+\dfrac{1}{2}a(\Delta t)^2\)

Arthur: \(x_f=0+(3~m/s)\Delta t+\dfrac{1}{2}(0)(\Delta t)^2=(3~m/s)\Delta t\)

Betty: \(x_f=100~m+(-2~m/s)\Delta t+\dfrac{1}{2}(0)(\Delta t)^2=100+(-2~m/s)\Delta t\)

Their final position, \(x_f\),is the same. Make the equations equal to each other:

\((3~m/s)\Delta t=100~m+(-2~m/s)\Delta t\)

Solve for \(\Delta t\): \(\Delta t=\dfrac{100~m}{2~m/s+3m/s}=\dfrac{100~m}{5~m/s}=20~s\)

Now we know the dog runs at 5 m/s, so if he ran for 20 s, he ran a total distance of:

\(\Delta x=v_o\Delta t=(5~m/s)*20s=100 ~m\)