Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differential equation dv m mg bv dt   . Here, m is the mass of the object, g is the acceleration due to gravity, and b > 0 is a constant. Consider an object that has a mass of 100 kilograms and an initial velocity of 10 m/sec (that is, v(0) = 10). If we take g to be 9.8 m/sec2 and b to be 5 kg/sec, find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

v = 196 - 186*e^( - 0.05*t )      

v-terminal = 196 m/s

Explanation:

Given:

- The differential equation for falling object velocity v in gravity with air resistance is given by:

                                m*dv/dt = m*g - b*v

- The initial conditions and constants are as follows:

                 v(0) = 10 , m = 100 kg , b = 5 kg/s , g = 9.8 m/s^2  

Find:

- Find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

Solution:

- Rewrite the differential equation in te form:

                                 dv/dt + (b/m)*v = g

- The integration factor function P(t) = b/m. The integrating factor u(t) is:

                                 u(t) = e^∫P(t).dt

                                 u(t) = e^∫(b/m).dt

                                 u(t) = e^[(b/m).t]

- Solve the differential equation after expressing in form:

                                 v.u(t) = ∫u(t).g.dt    

                                 v.e^[(b/m).t] = g*∫e^[(b/m).t].dt    

                                 v.e^[(b/m).t] = g*m*e^[(b/m).t] / b + C

                                  v = g*m/b + C*e^[-(b/m).t]                

- Apply the initial conditions v(0) = 10 m/s and evaluate C:

                                  10 = 9.8*100/5 + C*e^[-(b/m).0]

                                  10 = 9.8*100/5 + C

                                  C = -186

- The final ODE solution is:

                                  v = 196 - 186*e^( - 0.05*t )

- The Terminal velocity vt can be expressed by a limiting value for v(t), where t ->∞.        

                                  vt = Lim t ->∞ ( v(t) )

                                  vt = Lim t ->∞ ( 196 - 186*e^( - 0.05*t ) )

                                  vt = 196 - 0 = 196 m/s

a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be?

\

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length