1. 1. For velocity-dependent force, we can use the integral equation

      

      Here and F(v') = -bv' so the integral gives

      

      which we invert to obtain

      

      Integrating and applying the initial condition x(0) = 0 gives

      

      We can solve for in terms of x,

      

      to give v(x):

      

   2. As in review problem 2b, we can rearrange Newton's second law to directly relate v and x, by using the chain rule for the assumed solution v(x):

      

      This gives for Newton's second law, for velocity-dependent force,

      

      or

      

      (Note that the integrand really should be written in terms of dummy variables x', v', as position and velocity are varying up to the final values x and v.) Plugging in F = -bv gives

      

      since . This confirms our result in (a),

      

   3. Note that comes to rest in the finite distance . Once it comes to rest, the force -bv vanishes, so that the particle remains stopped.
2. We use the same coordinate system as in section 1.4; that is, x=0 at the earth's surface and increases upward, so that the skydiver feels force

   

   (gravity downward, air resistance up).

   The skydiver starts at t=0 at altitude x = h and velocity and falls, reaching altitude x and velocity v at time t. Thus at time t he has fallen through distance h-x, and we're asked to find velocity as a function of this distance fallen h-x.

   To do this we relate the integrand dx to a v-dependent integrand and integrate. As in problem 1, Newton's second law is rewritten in terms of v, x to give

   

   or

   

   With the limits above, and , we have

   

   or

   

   Let's call the distance fallen d, so d = h-x; we're asked to invert this relation to find v(d). Multiplying both sides by 2c/m then exponentiating gives

   

   Further rearranging to solve for v gives

   

   or

   

   noting the terminal velocity .

   This gives the general relation between velocity v and distance fallen d. Specifically, when

   

   Taking of both sides and rearranging,

   

   Note which here gives, since and , m/c = 297.6 m. Thus

   

3. 1. Note that the table's normal force opposes the force of gravity for the length l-x of cable which remains on the table. Gravity acts unopposed on the hanging length x. This portion has mass mx/l, so the total force pulling the cable over the table edge is F = mgx/l (acting to increase the hanging length x). Newton II thus gives

      

   2. The requested methods again use to rewrite Newton II as an equation relating v and x-dependent integrands:

      

      We integrate both sides for our motion, with F(x) = mgx/l and initial , v = 0. This gives (after cancelling factors of m)

      

   3. We know that dx/dt = v(x), so dt = dx/v(x), an equation between integrands each depending only on one variable, which we may integrate between the endpoints of our motion:

      

      Doing the integrals gives

      

      This is a rather ugly relationship, one not easily inverted to find x(t). It could be made a little bit nicer,

      

      but remains a bit baffling.

   4. We can confirm that , a solution found easily by other methods, obeys the initial conditions and our result of part c. First, initial conditions:

      

      do have as needed. Note that

      

      using the exponential definitions of cosh and sinh,

      

      This confirms our result of part c. Thus these closed integral methods remain consistent with ODE approaches, even when they're not very helpful.

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