Average velocities of runners?

Question

John and Mike both run with initial velocity of 9 m/s, until they completely lost their power (which, by the way, is proportional with square of velocity).

John's power is decreasing linearly by time, and Mike's power is decreasing linearly by distance.

Calculate average velocities for both of them!

(Sorry if my english is a little ununderstandable!)

Answer 1

Let's take John first:

v^2 = A - Bt.....(1)

Since v0=9 at t=0, eq (1) becomes:

v^2 = 81 - Bt..........(2)

giving total time T = 81/B.....(3)  (for v = 0 at the end).

By definition average speed is:

X = (1/T)*integral(from t=0 to T)[v*dt].........(4)

Differentiating (2) we get:

2v*dv = -B*dt.

Multiplying both sides by v and taking the integral you get:

2*integral(from v=v0 to 0)[v^2*dv] = -B*integral(t=0 to T)[v*dt],

which using (4) gives:

-B*T*X = 2*[v^3/3](from v0 to 0] = -2*v0^3/3 = -2*9^3/3.

Using (3) you get:

- 81*X = - 2*3*81, which gives John's average speed:

X = 6 m/s.

Similarly in Mike's case you get:

v^2 = 81 - Ds..........(5),

where s is the distance from the start. At the end v=0

giving total distance:

sT = 81/D..........(6).

Differentiating (5) you get:

2*v*dv = - D*v*dt, or:

2dv = -D*dt .........(7) (decelleration is constant).

Integrating (7) you get:

v = -(D/2)*t + K.

Plugging v0=9 for t=0 => K=9 giving:

v = -(D/2)*t + 9........(8).

At the end v=0 giving total travel time:

T = 18/D........(9).

Using the same average speed definition, eq. (4), integrating (8) you get:

T*X = [-(D/4)*t^2 + 9*t](from t=0 to T), which is:

T*X = -(D/4)*T^2 + 9*T, or:

X = -(D/4)*T + 9 = (using (9)) = 9 - 18/4 = 9/2 m/s.

  

Answer 2

1. m/s is a SPEED, not a "velosity"

2. power is proporshunal tu SPEED squared.

3. how long did em run at full power?

if em run full power 1 out & then take 2 sekonds tu die, get diff anser from if em run full speed 2 sekonds & take our tu fall down.

4. how long tu p**p out?

5. how far til em fall down?

2 many dont noes