Practice
practice problem 1
An alien is flying her spaceship at half the speed of light in the positive x direction
when the autopilot begins accelerating the ship uniformly in the negative y direction
at 2.34 m/s2 (0.2 times the acceleration due to gravity on the
alien's home planet, the name of which is impossible to write in human symbols). Determine the resultant
displacement and velocity of the spacecraft when the acceleration ceases 137 earth days later.
solution
| |
| Start your solution by listing the given quantities along each axis and the time in convenient units … |
| |
| t = 137 × 24 × 3600 = 1.184 × 107 s |
| |
| x0 = |
0 m |
y0 = |
0 m |
| vx0 = |
0.050(3 × 108 m/s) = 1.5 × 107 m/s |
vy0 = |
0 m/s |
| ax = |
0 m/s2 |
ay = |
−2.34 m/s2 |
| |
Apply the one-dimensional equations of motion for constant acceleration in each direction to get the
components of the displacement and velocity. |
| |
| x = |
x0 + vx0Δt + ½ axΔt2 |
y = |
y0 + vy0Δt + ½ ayΔt2 |
| x = |
vx0Δt |
y = |
½ ayΔt2 |
| x = |
(1.5 × 107 m/s)(1.184 × 107 s) |
y = |
½ (−2.34 m/s2)(1.184 × 107 s)2 |
| x = |
+1.776 × 1014 m |
y = |
−1.639 × 1014 m |
| |
| vx = |
vx0 + axΔt |
vy = |
vy0 + ayΔt |
| vx = |
vx0 |
vy = |
ayΔt |
| vx = |
1.5 × 107 m/s |
vy = |
(−2.34 m/s2)(1.184 × 107 s) |
| |
|
vy = |
−2.770 × 107 m/s |
| |
| Add the components to get the resultant displacement and velocity. |
| |
| r = |
√[ x2 + y2 ] |
v = |
√[ vx2 + vy2 ] |
| r = |
√[ (+1.776 × 1014 m)2 + (−1.639 × 1014 m)2 ] |
v = |
√[ (1.5 × 107 m/s)2 + (−2.770 × 107 m/s)2 ] |
| r = |
2.42 × 1014 m |
v = |
3.15 × 107 m/s |
| |
| tan θ = |
| y |
= |
−1.639 × 1014 m |
| x |
+1.776 × 1014 m |
|
tan θ = |
| vy |
= |
−2.770 × 107 m/s |
| vx |
1.5 × 107 m/s |
|
| θ = |
−42.7° |
θ = |
−61.6° |
| |
| r = |
2.42 × 1014 m at −42.7° |
v = |
3.15 × 107 m/s at −61.6° |
practice problem 2
An alien spacecraft accidentally flies into a plasma cloud (a collection of ionized
gas). This disrupts the ship's guidance system, which makes the velocity varying
according to the following parametric equations.
| |
|
| vx = 0.4 − cos (t/100) |
vy = 0.0 + sin (t/100) |
| |
|
To make the calculations simpler for us humans, the aliens have adapted this problem
to our standards. The equations use m/s for velocity, seconds for time, and radians
for angular measure. In addition, the initial coordinates of the ship were (0,0);
that is, the ship started acting this way when it was located at the origin.
Determine …
- the displacement as a function of time,
- the path of the ship for the first 2000 s,
- the direction of the ship at the beginning and end of this interval, and
- the maximum and minimum speed of the ship.
solution
- This is a problem that requires calculus. By definition, velocity is
the first derivative of displacement with respect to time. From the fundamental
theorem of calculus then, displacement is the time integral of velocity.
| |
|
| x = ∫ vx dt |
y = ∫ vy dt |
| x = ∫(0.4 − cos (t/100)) dt |
y = ∫(0.0 + sin (t/100)) dt |
| x = 0.4t − 100 sin (t/100) + cx |
y =− 100 cos (t/100) + cy |
| |
|
All that remains is to adjust the constants for the initial conditions:
x0 = y0 = 0.
| |
|
| 0 = |
0.4(0) − 100 sin (0/100) + cx |
0 = |
−100 cos (0/100) + cy |
| 0 = |
cx |
0 = |
−100 + cy |
| cx = |
0 |
cy = |
100 |
| |
|
Therefore
| |
|
| x = 0.4t − 100 sin (t/100) |
y = 100 − 100 cos (t/100) |
| |
|
- The ship is moving forward in the x direction, as the first term shows,
but also oscillating, as the second term shows. The coefficient of the
second term is much larger than the first, so we can be quite certain
that there will be times when the ship will reverse direction, but overall
the motion will be forward. It's sort of a "two steps forward, one
step back" kind of kind of shuffle.
At the same time, the ship is oscillating in the y direction. The
constant term is basically unimportant. The ship starts at a particular
location and then dodges side to side without ever making any headway
in this direction.
Combining these two motions results in a path that looks something like
a doodle from a bored physics student's notebook. Don't try to sketch
it. Let a computer or graphing calculator draw it for you.
- To determine the direction of motion, evaluate the velocity equations
at the appropriate times.
| |
|
| vx(t) = |
0.4 − cos (t/100) |
vy(t) = |
0.0 + sin (t/100) |
| vx(0) = |
0.4 − cos (0/100) |
vy(0) = |
0.0 + sin (0/100) |
| vx(0) = |
0.4 − 1.0 |
vy(0) = |
0.0 + 0.0 |
| vx(0) = |
− 0.6 m/s |
vy(0) = |
0.0 m/s |
| |
|
At the beginning of the interval, the velocity was entirely in the negative
x direction.
| |
|
| vx(t) = |
0.4 − cos (t/100) |
vy(t) = |
0.0 + sin (t/100) |
| vx(2000) = |
0.4 − cos (2000/100) |
vy(2000) = |
0.0 + sin (2000/100) |
| vx(2000) = |
0.4 − 0.4 |
vy(2000) = |
0.0 + 0.9 |
| vx(2000) = |
0.0 m/s |
vy(2000) = |
+ 0.9 m/s |
| |
|
At the end of the interval, the velocity was entirely in the positive
x direction.
- Speed is the magnitude of velocity and is found by adding the components
using the Pythagorean theorem.
| |
|
| v2 = |
(vx)2 + (vy)2 |
| v2 = |
(0.4 − cos (t/100))2 + (0.0 + sin (t/100))2 |
| v2 = |
0.16 − 0.8 cos (t/100) + cos2 (t/100) + sin2 (t/100) |
| v2 = |
0.16 − 0.8 cos (t/100) + 1.00 |
| v2 = |
1.16 − 0.8 cos (t/100) |
| |
|
The cosine term is what matters here. It has its maximum value of +1
when t/100 is 0, 2π, 4π and other even multiples of π and
its minimum value of −1 when t/100 is π,
3π, 5π and other odd multiples of π. To determine the
range of the ship's speed, simply substitute ±1 for the cosine
term.
| |
|
| v2 = |
1.16 − 0.8 cos (t/100) |
| vmin = |
√(1.16 −1.00) |
= √ 0.16 |
= 0.40 m/s |
| vmax = |
√(1.16 +1.00) |
= √ 2.16 |
= 1.47 m/s |
| |
|
practice problem 3
The parametric equations below are used for generating an interesting family of curves called
lissajous figures.
| |
|
| x = A sin (at + φ) |
y = B sin (bt) |
Where …
| x and y |
are coordinates on a plane |
| t |
is the parameter |
| A and B |
are amplitudes |
| a and b |
are angular frequencies |
| φ |
is the phase angle |
Use a graphing calculator or computer capable of graphing two-dimensional parametric equations. Set the
window dimensions to something like
| −1.5 < x < +1.5 |
−1.5 < y < +1.5 |
Since sine is a circular function, the range of parameter values should be
thought of in terms of laps around the unit circle.
We will use radians for all angular measures, so be sure your calculator
mode or computer preferences are set appropriately.
- Let the amplitudes and angular frequencies equal one (A = B = a = b = 1).
Set the parameter range to 0 < t < 2π with
a reasonable size increment (something that your calculator or computer
can complete in under thirty seconds). Draw the lissajous figures for
various phase angles. Try simple fractions of a complete circle like
φ = 0, ⅙π, ¼π, ½π, ⅔π, 1π, 1½π.
- What is the domain of x and y in all cases?
- What effect does phase angle have on the lissajous figure?
- Let the amplitudes equal one (A = B = 1).
Let the phase angle equal a quarter lap around the unit circle (φ = ½π).
Draw the lissajous figures for various angular frequencies.You will need to increase the maximum parameter
value to 4π, 6π, 20π or higher depending on your choice of frequencies.
- Try different whole number values to start with and do not choose
the same value for each frequency (a, b ∈ ℤ+ and a ≠ b).
Start with small numbers like 1, 2, 3, 4, 5.
Try larger numbers if you have the patience. Be sure to vary each frequency.
How is the appearance of the lissajous figure affected by your choice
of angular frequencies?
- Set one of the frequencies to 1 and the other to an irrational number like √2 or π. How is the behavior of this lissajous figure different from those of all your previous trials?
solution
practice problem 4
Write something completely different.
solution
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