Problem 2/29 Page 31, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: The force on a particle traveling along a straight line is linear with distance traveled.

Find: The maximum velocity

Relevant variables:

F(s) = m { K s + D }

m = Arrow mass

K s + D = amax    at s=0
K s + D = 0          at s=s0

amax = 16000 ft/s²
s0 = 2 ft

Thus:

D = amax

K = -amax/s0

1. Mechanical System: Arrow as it travels from s=0 to s=s0

2. Free Body Diagram : Shows arrow with bow force in direction of motion of arrow.

3. Equations:
   a. S F = m amax { 1 - s/s0 } t
   b. m a = m dv/ds ds/dt t = m dv/ds v t

4. Solving

m amax { 1 - s/s0 } = m v dv/ds

Dividing by m and separating variables:

v dv = amax { 1 - s/s0 } ds

Integrating both sides:

1/2 v² = amax { s - 1/2 s²/s0 } + E

Knowing that v=0 at s=0 permits the evaluation of E:
 
1/2 v² = amax s { 1 - 1/2 s/s0 }

The maximum velocity occurs when its derivative is equal to zero.  The derivative of velocity with respect to time is the acceleration.  Thus the maximum velocity occurs when the acceleration is zero, that is when s reaches s0.  Thus:

1/2 vmax² = amax s0 1/2

vmax = { amax s0 }^1/2 = 178.9 ft/s