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					IB Phys SL                                                                                   GOHS

2.5 Worksheet: Constant Acceleration and Kinematic Equations
Concept Questions:
10. Car A, traveling from New York to Miami, has a speed of 25 m/s. Car B, traveling from New
York to Chicago, also has a speed of 25 m/s. Are their velocities equal? Explain.

14. Consider the following combinations of signs and values for the velocity and        v  vo  at
acceleration of a particle with respect to a one-dimensional x-axis:
Velocity Acceleration                                                                          1
a. Positive Positive                                                                    x  (vo  v)t
b. Positive Negative                                                                           2
c. Positive Zero                                                                                    1
d. Negative Positive                                                                    x  vo t  at 2
e. Negative Negative                                                                                2
f. Negative Zero
g. Zero Positive                                                                        v 2  vo  2ax
                                                                                                2

h. Zero Negative
Describe what the particle is doing in each case, and give a real-life example for an automobile on
an east-west one-dimensional axis, with east considered the positive direction.

Problems:
25. In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-
m-long cannon with
final speed of 10.97 km/s. What would have been the unrealistically large acceleration
experienced by the space travelers during their launch? (A human can stand an acceleration of
15g for a short time.) Compare your answer with the free-fall acceleration, 9.80 m/s2.

31. A drag racer starts her car from rest and accelerates at 10.0 m/s2 for a distance of 400 m (
mile). (a) How long did it take the race car to travel this distance? (b) What is the speed of the
race car at the end of the run?

32. A jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of _5.00 m/s2
as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time
needed before it can come to rest? (b) Can this plane land on a small tropical island airport where
the runway is 0.800 km long?

33. A driver in a car traveling at a speed of 60 mi/h sees a deer 100 m away on the road. Calculate
the minimum constant acceleration that is necessary for the car to stop without hitting the deer
(assuming that the deer does not move in the meantime).

34. A record of travel along a straight path is as follows:
1. Start from rest with a constant acceleration of 2.77 m/s2 for 15.0 s.
2. Maintain a constant velocity for the next 2.05 min.
3. Apply a constant negative acceleration of _9.47 m/s2 for 4.39 s.
(a) What was the total displacement for the trip? (b) What were the average speeds for legs 1, 2,
and 3 of the trip, as well as for the complete trip?
ANSWERS:
Concept Questions:
10. Velocities are equal only if both magnitude and direction are the same. These
     objects are moving in different directions, so the velocities are not the same.
14.    (a) The car is moving to the east and increasing in speed.
       (b) The car is moving to the east but slowing in speed.
       (c) The car is moving to the east at constant speed.
       (d) The car is moving to the west but slowing in speed.
       (e) The car is moving to the west and speeding up.
       (f) The car is moving to the west at constant speed.
       (g) The car starts from rest and begins to speed up toward the east.
       (h) The car starts from rest and begins to speed up toward the west.

Problems:
2.25   From v2  v0  2a x , we have  10.  103 m s 2 0  2a 220 m
                  2
                                            97                                 so that
         a 2.  105 m s2 which is 2.  104 tm es g!
             74                     79       i


                           1 2                              1
2.31    (a) Using x  v0t at with v0  0 gives 400 m  0  10. m s2  t ,
                                                                0         2

                           2                                2

             yielding       94
                        t 8. s

        (b) From v  v0  at, with v0  0 , we find
             v 0  10. m s2   8. s  89. m s
                       0           94       4


                                              v  v0 0  100 m s
2.32    (a) The time required to stop is t                      20. s
                                                                     0
                                                a      5. m s2
                                                         00

        (b) The minimum distance needed to stop the plane is

                            v  v0   0  100 m s
                x  vavt          t            20. s  1000 m  1. km
                                                         0                00
                            2              2    

             Thus, the plane cannot stop in 0.8 km.
2.33   Using v2  v0  2a x , with v 0 and v0  60 m ih , yields
                   2




                     v2  v0 0   60 m ih    0.
                                            2
                                                           2
                           2
                                                   447 m s
                  a                                       3. m s
                                                                 6    2

                     2 x      2 100 m        1 m ih 


2.34   The velocity at the end of the first interval is

                                             0
                  v  v0  at 0  ( 77 m s) 15. s  41. m s
                                    2.                  6

       This is also the constant velocity during the second interval and the initial
       velocity for the third interval.

                         1 2
       (a) From x  v0t at , the total displacement is
                         2

             x   x1   x2   x3

                       1                   2
                   0   2. m s2  15. s    41. m s 123 s  0
                            77          0             6
                       2                                             

                                                             1                     2
                                         41. m s  4. s    9. m s2   4. s 
                                              6         39          47          39
                                                             2                      

             or    x  312 m  5.  103 m  91. m  5.  103 m  5. km
                                 11            2      51           51

                           x1 312 m
       (b)    vav 1                         20. m s
                                                   8
                           t1            0
                                       15. s

                           x2       5.  103 m
                                        11
              vav 2                            41. m s
                                                      6
                            t2           123 s

                           x3 91. m
                                   2
              vav 3                         20. m s , and the average velocity for the
                                                   8
                            t3           39
                                        4. s

                                             x        5.  103 m
                                                         51
             total trip is  vav total                             38. m s
                                                                         7
                                             total  15.
                                              t         0+123+4.  s
                                                                39