Who Was Ahead at the 6-Second Mark in the 100m Race?

[physics_geek]

Acellertaion and Velocity Help!

Setting a world record in a 100 m race, Maggie and Judy cross the finish line in a dead heat, both taking 10.0 s. Accelerating uniformly, Maggie took 2.00 s and Judy 2.80 s to attain maximum speed, which they maintained for the rest of the race.
(a) What was the acceleration of each sprinter?



(b) What were their respective maximum speeds?


(c) Which sprinter was ahead at the 6.00 s mark?
Maggie
Judy
neither


(d) By how much was that sprinter ahead?


ok..so I've talked to like 10 million people..and nobody seems to know how to do it..

i have no idea where to start...i have to find the acceleration first...im just not sure if their initial velocities would be 0 or what?

 

[Benzoate]

Setting a world record in a 100 m race, Maggie and Judy cross the finish line in a dead heat, both taking 10.0 s. Accelerating uniformly, Maggie took 2.00 s and Judy 2.80 s to attain maximum speed, which they maintained for the rest of the race.
(a) What was the acceleration of each sprinter?



(b) What were their respective maximum speeds?


(c) Which sprinter was ahead at the 6.00 s mark?
Maggie
Judy
neither


(d) By how much was that sprinter ahead?


ok..so I've talked to like 10 million people..and nobody seems to know how to do it..

i have no idea where to start...i have to find the acceleration first...im just not sure if their initial velocities would be 0 or what?

Well , since the acceleration is uniform , meaning constant, you would apply the kinematic equations for position, velocity and acceleration.

 

[physics_geek]

yea that's exactly what i thought!

but that's the thing..i don't know which equation to start with

i thought i would use the position equation
X final = Xinitial + Vinitial(t) + 1/2(a)t^2

and solve for acceleration

but i don't know what the initial velocity is..or how to find it

 

[DocZaius]

The way I approached this problem is to first consider one runner's graph of velocity in relation to time.

Since there's a uniform acceleration to a final velocity, the graph will have two phases. The first phase will be the acceleration phase which is a constant slope up to Vf (final velocity). The second phase will be a horizontal slope at V=Vf all the way until t=10.

The area under that graph is the distance covered and thus must be = 100.

For Maggie, you know the area of the rectangle in the second phase is Vf*8 (the number of seconds she spent at Vf). The first phase's area will be the area of a triangle because of the acceleration. That area will be (1/2)*Vf*(2).

Thus, (1/2)*Vf*2 + Vf*8 = 100
This gives you the final velocity. You can find accceleration from there, and so on.

 

[gutti]

Common sense,, the intial velocity is 0 saying as they haven't started running yet.

 

[nasu]

A slightly simpler solution:
Take one of the runners, for example the first one, and write the equation for the distance traveled (total):

d=1/2 a1 *t1^2 + vmax*(t-t1)
d=100m
t1=2s
vmax=a1*t1=a1*2s
t-t1=10s-2s=8s

Then you have
100m=0.5*a1*4+(a1*2s)*8s
Solve for a1;

Repeat for the other runner.

 

[physics_geek]

o thanks a lot

which sprinter was ahead at 6.00 s?

i think i use the position function...but i don't know what to plug in

 

[nasu]

Plug in the values for accelerations. Put t=6 seconds. Calculate the distance for each one.

 

[LowlyPion]

o thanks a lot

which sprinter was ahead at 6.00 s?

i think i use the position function...but i don't know what to plug in

You need to first determine the distance each sprinter covered for each of their acceleration phases.

x = 1/2*a*t²

Then you figure the additional distance covered at their top speed with the remaining time budget to 6 seconds.
As before you know their top speed was a*t of the acceleration phase.