Follow-up problems
3-39. An object's position as a function of time is
given by
r = (bt3 + ct)i + dt2j + (et + f) k, where b, c, d, e,
and f are constants. Determine the velocity and acceleration as
functions of time.
Solution:
We recall from our constants and equation sheet that
We can determine the velocity by taking the derivative of the given
vector function. This is done by taking the derivative of each
component independantly, or
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d
dt
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é
ë
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(bt3 + ct) |
^
i
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+ dt2 |
^
j
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+ (et + f) |
^
k
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|
ù
û
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d
dt
|
[bt3 + ct] |
^
i
|
+ |
d
dt
|
[dt2] |
^
j
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+ |
d
dt
|
[et + f] |
^
k
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(3bt2 + c) |
^
i
|
+ (2dt) |
^
j
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+ (e) |
^
k
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Now we must find the acceleration as a function of time. To
do this, we recall from our constants and equation sheet
Again, we take the derivative of each component independantly, giving us
a solution of
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d
dt
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é
ë
|
(3bt2 + c) |
^
i
|
+ (2dt) |
^
j
|
+ e |
^
k
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ù
û
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d
dt
|
[3bt2 + c] |
^
i
|
+ |
d
dt
|
[2dt] |
^
j
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+ |
d
dt
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[e] |
^
k
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3-43. A spacecraft is launched toward Mars at the
instant Earth is moving in the +x direction at its orbital speed of 30
km/s, in the Sun's frame of reference. Initially the spacecraft is
moving at 40 km/s relative to Earth, in the +y direction. At the
launch time, Mars is moving in the -y direction at its orbital speed
of 24 km/s. Find the spacecrafts velocity relative to Mars.
Solution:
We start by noting that the velocities of Earth and the
spacecraft add in the Sun's frame of reference. This allows us to write
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®
v
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s,sun
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= |
®
v
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s,Earth
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+ |
®
v
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Earth,sun
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This gives us the result
Now we must take into account the velocity of Mars relative to the Sun.
This is done by noting
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®
v
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s,Mars
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= |
®
v
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s,sun
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- |
®
v
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Mars,sun
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Using our result from above, we see this gives us
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[30 km/s |
^
x
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+ 40 km/s |
^
y
|
] - [-24 km/s |
^
y
|
] |
| |
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3-55. A ferryboat sails between two towns directly
opposite one another on a river. If the boat sails at 15 km/h relative
to the water, and if the current flows at 6.3 km/h, at what angle should
the boat head?
Solution:
We start the problem by determining our coordinate system. We
will let the river flow in the i-direction (
[(v)\vec]riv = vriv i), and have the towns
lie such that travel in the j-direction will result in
moving from one town to the other. So we want
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®
v
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net
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= |
®
v
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boat
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+ |
®
v
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riv
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Since we want [(v)\vec]net to have no
[^(i)]-component, we can see that we can solve the equation
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cos-1 |
æ
ç
è
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- |
vriv
vboat
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ö
÷
ø
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cos-1 |
æ
ç
è
|
- |
6.3 km/h
15 km/h
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ö
÷
ø
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Solutions translated from TEX by TTH, version 1.57.
