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Recitation 2
(Chapter 2) Jan. 2428
Ch. 2 # 9.
A tourist being chased by an angry bear is running in a
straight line toward his car at a speed of 4.0 m/s. The car is a distance
d
away. The bear is 26 m behind the tourist and running at 6.0 m/s. The
tourist reaches the car safely. What is the maximum possible value for
d
?
REASONING
In order for the bear to catch the tourist over the distance
d
, the bear must reach the car at the same time as the tourist.
During the time
t
that
it takes for the tourist to reach the car, the bear must travel a total distance of
d
+ 26
m.
From Equation 2.1,
tourist
d
v
t
=
(1)
and
bear
26 m
d
v
t
+
=
(2)
Equations (1) and (2) can be solved simultaneously to find
d
.
SOLUTION
Solving Equation (1) for
t
and substituting into Equation (2), we find
tourist
bear
tourist
(
26 m)
26 m
/
d
v
d
v
d v
d
+
+
=
=
bear
tourist
26 m
1
v
v
d
=
+
Solving for
d
yields:
bear
tourist
26 m
26 m
52 m
6.0 m/s
1
1
4.0 m/s
d
v
v
=
=
=


______________________________________________________________________________
Ch. 2 # 18.
A sprinter explodes out of the starting block with an acceleration of +2.3 m/s
2
, which she sustains
for 1.2 s. Then, her acceleration drops to zero for the rest of the race.
(a) What is her velocity at
t
= 1.2 s.
(b) What is her velocity at the end of the race?
REASONING
We can use the definition of average acceleration
(
29
(
29
0
/
t
t
=


0
a
v
v
(Equation 2.4) to find
the sprinter’s final velocity
v
at the end of the acceleration phase, because her initial velocity (
0 m/s
=
0
v
,
since she starts from rest), her average acceleration
a
, and the time interval
t

t
0
are known.
SOLUTION
a. Since the sprinter has a constant acceleration, it is also equal to her average acceleration, so
2
2.3 m/s
= +
a
Her velocity at the end of the 1.2s period is
(
29
(
29
(
29
(
29
2
0
0 m/s
2.3 m/s
1.2 s
2.8 m/s
t
t
=
+

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 Spring '09
 Staff

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