Solution Set to Assignment 4

 

Some notes to explain the equations:

 

§      The symbol D means “change in” so that DT is the change in T (whatever T represents).  The change in something is the difference between what it is at the end, or finally, T_f and what it was at the start, or initially, T_i  DT = T_f - T_i

 

§      I can’t do vector arrows on the computer so I will use another notation that is common.  Bold symbols represent vectors, so p is a vector quantity while p is a scalar quantity.

 

§      x  represents the position of an object and t represents a time.

 

Equations:

 

1.    Elapsed time = Dt = t_f - t_i

2.    Displacement = Dx = x_f - x_i

3.    Velocity = v = Dx /Dt

4.    Acceleration =  a = Dv /Dt  (where, Dv = v_f - v_i)

 

In the case of free fall under the influence of gravity there are two equations.  To make it easier on you I will consider only cases where objects fall from rest (that is, they have NO initial velocity).  Down is considered positive here and objects start from position zero at time zero. The (constant) acceleration due to gravity is given by g = 9.8m/s²

 

5.     x = 1/2gt²

6.     v = gt

 

 

 

Q1

a)    Use equation 1 with t_f = 3:17:30 and  t_i = 3:15:15.  This gives Dt=2:15, or better, 135 s.

b)   Use equation 2 with  x_f = 1km West and x_i = 0km.  This gives Dx = 1km West, or better, 1000m West.

c)   Use equation 3 with the answers above to get v = 1000mWest/135s = 7.407m/s West.

d)   Speed is velocity with no direction when traveling always in same direction, so speed = 7.407m/s.

 

Q2

a)    200 min is 3:20:00, so added to 1:00:00 gives 4:20:00.

b)   Use equation 3 but solve for displacement to get Dx = vDt, and use = v=5.0m/s North and Dt = 200 min which in seconds is 12,000s.  This gives Dx = 5.0m/s North ´ 12,000s = 60,000m North.

c)   Use equation 2 but solve it for final position, x_f (which is what the question is asking for).  This gives  x_f = x_i + Dx  and you know  Dx  = from answer b) and x_i = 500 m South from the question.  If you take North as positive direction you can calculate  x_f = -500m + 60,000m  = +59,500m which means 59,500m North of Oruro.

d)   Acceleration is given in equation 4.  Since we have a constant velocity here, Dv = 0 so a = 0.

 

Q3

     a) Use equation 4 with (we will choose up the Pando as the positive direction), v_i= +10m/s and  

         v_f = +15m/s, and Dt = 5.0s.  This gives  Dv = +5m/s and so a = +5m/s / 5.0s = +1.0 m/s², or better                      

        1.0m/s² up the Pando (North)

     b) A similar argument is used in the braking process, but now v_i= +15m/s and  

         v_f = 0m/s, so Dv = -15m/s  and Dt = 3.0s so so a = -15m/s / 3.0s = -5.0m/s², or better                      

        5.0m/s² down the Pando (South).

 

Q4

a)    Use equation 5 with t=0.05s and get x = +0.01225 m OR 1.225 cm.  That is, 1.225 cm DOWN

     from the table top level.

b)   Use equation 6 with time as above and get v = 0.49m/s DOWN.

c)   Now you have to use equation 5, but solve it for time, getting t = Ö(2x/g).  Using the fact that x = +1.0m when the object hits the floor (since the table is 1.0 m high, and down is positive and the table top can be called zero position.), we get t = Ö(2(1.0m)/9.8m/ s²) = 0.45s.  (Check the units algebraically and you will see that they work out to seconds.)

d)   You can now use equation 6 again with the time found in answer c) to find the speed with which it hits the floor.  You get v = 4.43 m/s DOWN, so the speed is 4.43m/s.