# Relationship Between Velocities of Two Runners

1. May 30, 2015

### Drakkith

Staff Emeritus
1. The problem statement, all variables and given/known data
Reginald is out for a morning jog, and during the course of his run on a straight track, he has a velocity that depends upon time as shown in the figure below. That is, he begins at rest, and ends at rest, peaking at a maximum velocity Vmax at an arbitrary time tmax. A second runner, Josie, runs throughout the time interval t = 0 to t = t f at a constant speed Vj, so that each has the same displacement during the time interval. Note: t f is NOT twice t max , but represents an arbitrary time. What is relation between Vj and tmax?

2. Relevant equations
Position equation for constant acceleration: X=Xi+Vi+1/2AT2
Velocity Equation for constant acceleration: V=Vi+AT
Position Equation for constant velocity: X=Xi+VT

3. The attempt at a solution

Not sure what to do really, so I've just been trying things.
The only commonality between the two runners is the final position X and the time tf

For Josie, X = Vjtf

For Reginald:
Velocity:
From ti to tmax, Vmax = Atmax
From tmax to tf, since Vf = 0, the equation is: 0 = Vmax + Atf

Position:
From ti to tmax, initial displacement and velocity are zero: X1 = 1/2Atmax2
From tmax to tf: X = X1 + Vmaxtf + 1/2Atf2

Since X = Vjtf, we can rewrite the above equation as: Vjtf = X1 + Vmaxtf + Atf2
Replacing X1 with its equation: Vjtf = 1/2Atmax2+ Vmaxtf + Atf2

That's about as far as I've gotten and I don't know if I'm even on the right track.

2. May 30, 2015

### jbriggs444

Distance travelled is the integral of velocity over time -- the area under the velocity versus time graph. Given that, the problem turns into a simple geometric exercise. One needn't bother with any equations at all.

3. May 30, 2015

### Drakkith

Staff Emeritus
Thanks Jbriggs. Apparently the question was asking about Vmax, not Tmax, as Vj = 1/2 Vmax is the correct answer. (Which wouldn't have helped me prior to your post anyways, I still had no idea to look for the area under the graph)

Edit: Just so I don't look like an idiot, the question I posted was a literal copy and paste. I didn't just misread Vmax as Tmax.

Last edited: May 30, 2015