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Unformatted text preview: traveling at 89.5 k m h . Thus, the distance traveled is
x = v t = ( 89.5 k m h ) t1 = ( 77.8 k m h ) ( t1 + 0.367 h ) or, ( 89.5 km h ) t1 = (77.8 km h ) t1 + 28.5 km . From which, t1 = 2.44 h for a total time of t = t1 + 0.367 h = 2.80 h .
Therefore, 2.12 x = v t = ( 77.8 km h ) ( 2.80 h ) = 218 km . (a) At the end of the race, the tortoise has been moving for time t and the hare for a
time t − 2 .0 m in = t − 120 s . The speed of the tortoise is v t = 0.100 m s , and the speed
of the hare is v h = 20 v t = 2 .0 m s . The tortoise travels distance x t , which is 0.20 m
larger than the distance x h traveled by the hare. Hence, x t = x h + 0.20 m , which
becomes v t t = v h ( t − 120 s ) + 0.20 m or ( 0.100 m s ) t = ( 2 .0 m s ) ( t − 120 s ) + 0.20 m . This gives the time of the race as t = 1.3 × 102 s . ( ) (b) x t = v t t = ( 0.100 m s ) 1.3 × 102 s = 13 m 24 CHAPTER 2.13 2 The maximum time to complete the trip is total d ist an ce
1600 m ⎛ 1 km h ⎞
= 23.0 s .
=
r eq u ir ed a v er a ge sp eed 250 km h ⎜ 0.278 m s ⎟
⎝
⎠ tt = The time spent in the first half of the trip is h a lf d ist a n ce
800 m ⎛ 1 km h ⎞
=
= 12 .5 s .
v1
230 km h ⎜ 0.278 m s ⎟
⎝
⎠ t1 = Thus, the maximum time that can be spent on the second half of the trip is
t 2 = t t t1 = 23.0 s − 12 .5 s = 10.5 s , and the required average speed on the second half is v2 = 2.14 ⎛ 1 km h ⎞
h a lf d ist a n ce 800 m
= 76.2 m s ⎜
= 274 k m h .
=
t2
10.5 s
⎝ 0.278 m s ⎟
⎠ Choose a coordinate axis with the origin at the flagpole and east as the positive
direction. Then, using x = x i + v i t + 1 at 2 with a = 0 for each runner, the xcoordinate of
2
each runner at time t is
x A = − 4.0 m i + ( 6.0 m i h ) t x B = 3.0 m i + ( − 5.0 m i h ) t an d When the runners meet, x A = x B
or − 4.0 m i + ( 6.0 m i h ) t = 3.0 m i + ( − 5.0 m i h ) t . 7.0 m i
= 0.64 h . At this time,
11.0 m i h
x A = x B = −0.18 m i . Thus, they meet 0.18 m i w e st o f t h e fla g p o le ....
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 Spring '05
 Dr.Ha
 Physics, Acceleration

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