Physics of Sports, volume 3

Kinematics

Answers to Questions.

last updated on 21 April/98

Question set 3A.

1. Throwing a pitch on the train. What does the ground observer see. Let the train go in the positive x direction.

with the train, the v = vthrow + v train = 40 + 50 = 90 m/s
against the train. v= vthrow - vtrain = -40 +50 = 10m/s
transverse to the train. This is a little more complicated and requires 2-D vector addition. Think of the train going in the x direction and the ball thrown in the y direction. The net velocity seen by the ground observer is the vector sum so the ball is going in the x and y directions. The net velocity is the hypotenuse of the t riangle v = sqrt(40^2+50^2) = 64m/s.

2. Pitching a distance of 18m with v = 40m/s .. v = x/t ; t = x/v = 18/40 sec = .45 s

3. The relative speed of the pitched ball was 130m/s from the observer going 100m/s in the same direction as the ball. This means the ball was going VERY fast. For the ground observer, the ball must be going v = vrel + vplane = 130 +100 =230m/s

t = x/v = 18/230 sec = 0.078 sec!!

4. Swimmer can swim at 2m/s. The river is flowing at 5m/s.

If the swimmer aims herself straight across, she will cross the 100m wide river most quickly by maximizing the cross river velocity. t = x/v = 100/2 = 50sec. She would end up downstream by y =v*t = 5*50 = 250 m.
If the swimmer swims the shortest distance, she must aim upriver to try to effectively swim straight across. Unfortunately, the river is flowing too fast! Even if she swims straight upriver, she would still be carried downstream at 3m/s and never get across in any case. Let's try a test case, for aiming herself 45 degrees upstream. Her cross river velocity is reduced to be 2/sqrt(2)= 1.41 m/s [This is geometry]. she would take t = 100/1.41= 71 seconds to cross. Her upstream velocity is 1.41m/s relative to the water so her velocity relative to the shore is 5 - 1.41 = 3.6m/s. during the 71 seconds crossing time, she would be carried downstream by 255m! This is even farther downstream than the first case!

Question set 3B.

1. Carl Lewis and Racing. From the description, it appears that Lewis has less acceleration than other racers since he falls behind in the first part of the race. for the entire race he finishes first so he has the highest average speed. Lewis has superior speed but lesser acceleration compared with other sprinters. In the 4 x 100 race, when Lewis has a running start, he really shines!!

2. Kaufmann, the RB has it all... high acceleration and high speed. When he is carrying the ball in the open field, he can accelerate away from most tacklers when he meets them one on one and have the tacklers grasping on air.

3. Acceleration of Lewis. a = dv/dt = 11/4 = 2.8 m/s/s

Question set 3C.

1.1 converting distance records from English to International Units.

29' 2.5" = 29.21'*1m/3.3' = 8.9 m
18' 5.75" = 18.48' * 1m/3.3' =5.6 m
224'5" = 224.4'*1m/3.3'=68m

1.2 converting time.

8min + 13.8 s = 8*60+13.8 = 493.8 sec
2 hr 10 min = 2*60*60 + 10 * 60 = 7800s
1 week * 7days/week * 24hrs/day *3600sec/hr = 604,800 sec
1 day * 24hr/day*3600sec/hr = 86400 sec
1 yr ~ pi * E7 sec.

1.3 55 mi/hr

55mi/hr * 1.6km/mi = 88 km/hr
55mi/hr * (1m/s)/(2.2 mi/hr)= 25m/s

1.4 Marathon

v = d/t = (26mi + 384 yd)/2:20
Here we must use 384 yd* 1mi/1760 yd = .22 mi
distance = 26.22mi * 1600m/1mi =42000m
time = 2:20 = 2*3600 + 20*60 = 8400 sec
v = 42000/8400 = 5 m/s [Half the speed of a sprinter]

1.5 velocity from the table.

x	t	v
0	0
2.7	1	2.7/1 = 2.7m/s
21.6	2	18.9/1 = 18.9m/s
72.9	3	51.3/1 = 51.3m/s
172.8	4	99.9/1 = 99.9m/s

1.6 distance time plot.

At 0.5 sec, the velocity is.8 m/s at at 1.5 sec, it's about 2 m/s

1.7 distance plots for various conditions of velocity.

1.8

1.9 splits for a 1 mile race.

distance(yards)	time	time (sec)	Dx(m)	Dt(s)	V(m/s)
0	0	0
440	0:59	59	400	59	6.78
880	1:57	117	400	58	6.9
1320	3:03	183	400	66	6.06
1760	4:02	242	400	59	6.78

overall velocity = 1600/242 = 6.6 m/s

1.10 1500 m race

A: t =x/v = 1500/7.3 = 205 s
B t = x/v = 1500/7.4 = 203 s
difference in time = 2 seconds.

1.11 Late starter in the 1500m race.

In this problem, if B starts 3 seconds later, he will never catch A before the finish line. From the above calculation, if B starts 2 seconds behind, he will just catch A at the finish. With a longer time, no catching is possible.

1.12 the pitch.

a) 100 mi/hr * 1m/s/2.2mi/hr = 45m/s. t = x/v = 18/45 = 0.4 sec

b) 80 mi/hr /2.2 = 36m/s. t = x/t = 18/36 = 0.5 sec