2.9.
Solve:
(a)
The time for each of segment is
and
The average speed to the house is
1
50 mi/40 mph
5/4 hr
t
∆=
=
2
50 mi/60 mph
5/6hr.
t
=
100 mi
48 mph
5/6 hr
5/4 hr
=
+
(b)
Julie drives the distance
∆
x
1
in time
∆
t
1
at 40 mph. She then drives the distance
∆
x
2
in time
∆
t
2
at 60 mph. She
spends the same amount of time at each speed, thus
12
1
2
1
/40 mph
/60 mph
(2/3)
tt x
x
x
∆=∆⇒∆
=∆
⇒∆ =
∆
2
x
But
∆
x
1
+
∆
x
2
=
100 miles, so (2/3)
∆
x
2
+
∆
x
2
=
100 miles. This means
∆
x
2
=
60 miles and
∆
x
1
=
40 miles. Thus, the
times spent at each speed are
∆
t
1
=
40 mi/40 mph
=
1.00 hr and
∆
t
2
=
60 mi/60 mph
=
1.00 hr. The total time for her
return trip is
∆
t
1
+
∆
t
2
=
2.00 hr. So, her average speed is 100 mi/2 hr
=
50 mph.
2.16.
Model:
Represent the jet plane as a particle
Visualize:
Solve:
(a)
Since we don’t know the time of acceleration, we will use
22
10
2
2
2
1
2(
)
(400 m/s)
(300 m/s)
8.75 m/s
2
2(4000 m)
vv a
xx
vv
a
x
=+
−
−
−
⇒=
=
=
(b)
The acceleration of the jet is approximately equal to
g
,
the acceleration due to gravity.
2.22.
Model:
We will use the particle model and the constantacceleration kinematic equations.
Visualize:
Solve:
(a)
Substituting the known values into
2
1
100
2
yyvt a
t
=
+∆
+ ∆
, we get
11
1
10 m
0 m
20 (m/s)
( 9.8 m/s )
2
tt
−=+
+
−
One of the roots of this equation is negative and is not relevant physically. The other root is
t
1
=
4.53 s which is the
answer to part (b). Using
, we obtain
vva
=+∆
t
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1
20(m/s)
( 9.8 m/s )(4.53 s)
24.4 m/s
v
=+
−
=
−
(b)
The time is 4.53 s.
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 Spring '06
 staff
 Acceleration, Velocity, m/s, Tina

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