Runner A is initially 4 mi west of a flagpole and is running with a constant velocity of 6 mi/h due east...?

Question

and RunnerB is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5 mi/h due west...

How far are the runners from the flagpole when their paths cross?

Answer 1

NOTE: Xfa is the final location of runner A, Xa is the initial location of runner A, Xfb is the final location of runner B, Xb is the initial location of runner B, t is time, V0 is the initial velocity or in this case the constant velocity, ^ means there is an exponent (example t^2 is time squared), and a is the acceleration which in this case is equal to zero because the velocity is constant.

NOTE: The west direction is negative and the east direction is positive on the x axis.  This means that runner A is starting from a negative point in respect to the flag pole which is at x=0.  This also means that Runner B’s velocity is negative because he is running in the west direction which is negative.

GIVEN for Runner A:  Xa= -4 miles, V0= 6 mi/h, a= 0

GIVEN for Runner B:  Xb= 3 miles, V0= -5 mi/h, a=0

Find:  Xfa and Xfb

Xfa=Xfb

Xfa=Xa+v0t-(1/2)(a)(t^2)

Xfb=Xb+v0t+(1/2)(a)(t^2)

Xa+v0t-(1/2)(a)(t^2)= Xb+v0t+(1/2)(a)(t^2)

-4 mi + (6 mi/h)(t)+(1/2)(0 mi/h^2)(t)=3 mi+( - 5 mi/h)(t)-(1/2)( 0 mi/h^2)(t)

- 4 mi + 6 mi/h(t)-0=3 mi – 5 mi/h(t)-0

-4 mi + 6 mi/h(t) = 3 mi – 5 mi/h(t)

6 mi/h(t) = 7 mi – 5 mi/h(t)

11 mi/h(t) = 7 mi

t = (7 mi)  / (11 mi/h)

Xfa=Xa+v0t-(1/2)(a)(t^2)

Xfa=Xa+v0t

Xfa= - 4 mi + 6 mi/h((7 mi) / (11 mi/h))

Xfa= - 0.18 mi

CHECK:

Xfb=Xb+v0t+(1/2)(a)(t^2)

Xfb=Xa+v0t

Xfa= 3 - 5 mi/h((7 mi) / (11 mi/h))

Xfb= - 0.18 mi

So, the runners will meet at 0.18 miles west of the flag pole.