Motion in Space

Practice (40)

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1. A particle moves in a parabolic path defined by the vector-valued function \(\text{r}(t)={t}^{2}\text{i}+\sqrt{5-{t}^{2}}\text{j},\) where t measures time in seconds.

   1. Find the velocity, acceleration, and speed as functions of time.
   2. Sketch the curve along with the velocity vector at time \(t=1.\)
   Rivela la risposta

   1. We use , , and :
      \[\begin{array}{lll}\text{v}(t) & = & {r}^{'}(t)=2t\ \text{i}-\frac{t}{\sqrt{5-{t}^{2}}}\text{j} \\ \text{a}(t) & = & {v}^{'}(t)=2\text{i}-5{(5-{t}^{2})}^{-\frac{3}{2}}\text{j} \\ v(t) & = & ‖{r}^{'}(t)‖ \\ & = & \sqrt{{(2t)}^{2}+{(-\frac{t}{\sqrt{5-{t}^{2}}})}^{2}} \\ & = & \sqrt{4{t}^{2}+\frac{{t}^{2}}{5-{t}^{2}}} \\ & = & \sqrt{\frac{21{t}^{2}-4{t}^{4}}{5-{t}^{2}}}.\end{array}\]
   2. The graph of \(\text{r}(t)={t}^{2}\text{i}+\sqrt{5-{t}^{2}}\text{j}\) is a portion of a parabola (). The velocity vector at \(t=1\) is
      \[\text{v}(1)={r}^{'}(1)=2(1)\ \text{i}-\frac{1}{\sqrt{5-{(1)}^{2}}}\text{j}=2\text{i}-\frac{1}{2}\text{j}\]
      and the acceleration vector at \(t=1\) is
      \[\text{a}(1)={v}^{'}(1)=2\text{i}-5{(5-{(1)}^{2})}^{\text{-}3\text{/}2}\text{j}=2\text{i}-\frac{5}{8}\text{j}.\]
      Notice that the velocity vector is tangent to the path, as is always the case.
      

2. A particle moves in a path defined by the vector-valued function \(\text{r}(t)=({t}^{2}-3t)\ \text{i}+(2t-4)\ \text{j}+(t+2)\text{k},\) where t measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.

   Rivela la risposta

   \(\begin{array}{l}\text{v}(t)={r}^{'}(t)=(2t-3)\ \text{i}+2\text{j}+\text{k} \\ \text{a}(t)={v}^{'}(t)=2\text{i} \\ v(t)=‖{r}^{'}(t)‖=\sqrt{{(2t-3)}^{2}+{2}^{2}+{1}^{2}}=\sqrt{4{t}^{2}-12t+14}\end{array}\)

   The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.

3. A particle moves in a path defined by the vector-valued function \(\text{r}(t)={t}^{2}\text{i}+(2t-3)\ \text{j}+(3{t}^{2}-3t)\ \text{k},\) where t measures time in seconds and distance is measured in feet.

   1. Find \({a}_{\text{T}}\) and \({a}_{\text{N}}\) as functions of t.
   2. Find \({a}_{\text{T}}\) and \({a}_{\text{N}}\) at time \(t=2.\)
   Rivela la risposta

   1. Let’s start with :
      \[\begin{array}{lll}\text{v}(t) & = & {r}^{'}(t)=2t\ \text{i}+2\text{j}+(6t-3)\ \text{k} \\ \text{a}(t) & = & {v}^{'}(t)=2\text{i}+6\text{k} \\ {a}_{\text{T}} & = & \frac{\text{v}\cdot \text{a}}{‖\text{v}‖} \\ & = & \frac{(2t\ \text{i}+2\text{j}+(6t-3)\ \text{k})\cdot (2\text{i}+6\text{k})}{‖2t\ \text{i}+2\text{j}+(6t-3)\ \text{k}‖} \\ & = & \frac{4t+6(6t-3)}{\sqrt{{(2t)}^{2}+{2}^{2}+{(6t-3)}^{2}}} \\ & = & \frac{40t-18}{\sqrt{40{t}^{2}-36t+13}}.\end{array}\]
      Then we apply :
      \[\begin{array}{ll}{a}_{\text{N}} & =\sqrt{{‖\text{a}‖}^{2}-{a\ \text{T}}^{2}} \\ & =\sqrt{{‖2\text{i}+6\text{k}‖}^{2}-{(\frac{40t-18}{\sqrt{40{t}^{2}-36t+13}})}^{2}} \\ & =\sqrt{4+36-\frac{{(40t-18)}^{2}}{40{t}^{2}-36t+13}} \\ & =\sqrt{\frac{40(40{t}^{2}-36t+13)-(1600{t}^{2}-1440t+324)}{40{t}^{2}-36t+13}} \\ & =\sqrt{\frac{196}{40{t}^{2}-36t+13}} \\ & =\frac{14}{\sqrt{40{t}^{2}-36t+13}}.\end{array}\]
   2. We must evaluate each of the answers from part a. at \(t=2\text{:}\)
      \[\begin{array}{lll}{a}_{\text{T}}(2) & = & \frac{40(2)-18}{\sqrt{40{(2)}^{2}-36(2)+13}} \\ & = & \frac{80-18}{\sqrt{160-72+13}}=\frac{62}{\sqrt{101}} \\ {a}_{\text{N}}(2) & = & \frac{14}{\sqrt{40{(2)}^{2}-36(2)+13}} \\ & = & \frac{14}{\sqrt{160-72+13}}=\frac{14}{\sqrt{101}}.\end{array}\]
      The units of acceleration are feet per second squared, as are the units of the normal and tangential components of acceleration.

4. An object moves in a path defined by the vector-valued function \(\text{r}(t)=4t\ \text{i}+{t}^{2}\text{j},\) where t measures time in seconds.

   1. Find \({a}_{\text{T}}\) and \({a}_{\text{N}}\) as functions of t.
   2. Find \({a}_{\text{T}}\) and \({a}_{\text{N}}\) at time \(t=-3.\)
   Rivela la risposta

   1. \(\begin{array}{lll}\text{v}(t) & = & {r}^{'}(t)=4\text{i}+2t\ \text{j} \\ \text{a}(t) & = & {v}^{'}(t)=2\text{j} \\ {a}_{\text{T}} & = & \frac{2t}{\sqrt{{t}^{2}+4}},{a}_{\text{N}}=\frac{4}{\sqrt{4+{t}^{2}}}\end{array}\)
   2. \({a}_{\text{T}}(-3)=-\frac{6\sqrt{13}}{13},{a}_{\text{N}}(-3)=\frac{2\sqrt{13}}{13}\)

5. During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. The cannon is aimed at an angle of 30° above horizontal and the initial speed of the cannonball is \(600\ \text{ft/sec}\text{.}\) The cliff is 100 ft above the water ().

   1. Find the maximum height of the cannonball.
   2. How long will it take for the cannonball to splash into the sea?
   3. How far out to sea will the cannonball hit the water?
      
   Rivela la risposta

   We use the equation

   \[\text{s}(t)={v}_{0}t\ \text{cos}\ \theta \ \text{i}+({v}_{0}t\ \text{sin}\ \theta -\frac{1}{2}g{t}^{2})\ \text{j}\]

   with \(\theta =30\text{^{\circ}},\) \(g=32{\ \text{ft/sec}}^{2},\) and \({v}_{0}=600\) ft/sec. Then the position equation becomes

   \[\begin{array}{ll}\text{s}(t) & =600t(\text{cos}\ 30)\ \text{i}+(600t\ \text{sin}\ 30-\frac{1}{2}(32){t}^{2})\ \text{j} \\ & =300t\sqrt{3}\text{i}+(300t-16{t}^{2})\ \text{j}.\end{array}\]

   1. The cannonball reaches its maximum height when the vertical component of its velocity is zero, because the cannonball is neither rising nor falling at that point. The velocity vector is
      \[\begin{array}{ll}\text{v}(t) & ={s}^{'}(t) \\ & =300\sqrt{3}\text{i}+(300-32t)\ \text{j}.\end{array}\]
      Therefore, the vertical component of velocity is given by the expression \(300-32t.\) Setting this expression equal to zero and solving for t gives \(t=9.375\) sec. The height of the cannonball at this time is given by the vertical component of the position vector, evaluated at \(t=9.375.\)
      \[\begin{array}{ll}\text{s}(9.375) & =300(9.375)\sqrt{3}\text{i}+(300(9.375)-16{(9.375)}^{2})\ \text{j} \\ & =4871.39\text{i}+1406.25\text{j}\end{array}\]
      Therefore, the maximum height of the cannonball is 1406.39 ft above the cannon, or 1506.39 ft above sea level.
   2. When the cannonball lands in the water, it is 100 ft below the cannon. Therefore, the vertical component of the position vector is equal to \(-100.\) Setting the vertical component of \(\text{s}(t)\) equal to \(-100\) and solving, we obtain
      \[\begin{array}{lll} \\ 300t-16{t}^{2} & = & -100 \\ 16{t}^{2}-300t-100 & = & 0 \\ 4{t}^{2}-75t-25 & = & 0 \\ t & = & \frac{75\pm \sqrt{{(-75)}^{2}-4(4)(-25)}}{2(4)} \\ & = & \frac{75\pm \sqrt{6025}}{8} \\ & = & \frac{75\pm 5\sqrt{241}}{8}.\end{array}\]
      The positive value of t that solves this equation is approximately 19.08. Therefore, the cannonball hits the water after approximately 19.08 sec.
   3. To find the distance out to sea, we simply substitute the answer from part (b) into \(\text{s}(t)\text{:}\)
      \[\begin{array}{ll}\text{s}(19.08) & =300(19.08)\sqrt{3}\text{i}+(300(19.08)-16{(19.08)}^{2})\ \text{j} \\ & =9914.26\ \text{i}-100.7424\ \text{j}.\end{array}\]
      Therefore, the ball hits the water about 9914.26 ft away from the base of the cliff. Notice that the vertical component of the position vector is very close to \(-100,\) which tells us that the ball just hit the water. Note that 9914.26 feet is not the true range of the cannon since the cannonball lands in the ocean at a location below the cannon. The range of the cannon would be determined by finding how far out the cannonball is when its height is 100 ft above the water (the same as the altitude of the cannon).

6. An archer fires an arrow at an angle of 40° above the horizontal with an initial speed of 98 m/sec. The height of the archer is 171.5 cm. Find the horizontal distance the arrow travels before it hits the ground.

   Rivela la risposta

   967.15 m

7. Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object other than the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes

   \[{P}^{2}=\frac{4{\pi }^{2}{a}^{3}}{G(m+M)},\]

   where m is the mass of the Moon and M is the mass of Earth, a represents the length of the major axis of the elliptical orbit, and P represents the period.

   Given that the mass of the Moon is \(7.35\ \times \ {10}^{22}\ \text{kg,}\) the mass of Earth is \(5.97\ \times \ {10}^{24}\ \text{kg,}\) \(G=6.67\ \times \ {10}^{-11}{\text{m}}^{3}\text{/}\text{kg}\cdot {\text{sec}}^{2},\) and the period of the moon is 27.3 days, let’s find the length of the major axis of the orbit of the Moon around Earth.

   Rivela la risposta

   It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units, we need to use seconds for the period of the Moon as well:

   \[27.3\ \text{days}\ \times \ \frac{24\ \text{hr}}{1\ \text{day}}\ \times \ \frac{3600\ \text{sec}}{1\ \text{hour}}=2,358,720\ \text{sec}\text{.}\]

   Substitute all the data into and solve for a:

   \[\begin{array}{lll}{(2,358,720\ \text{sec})}^{2} & = & \frac{4{\pi }^{2}{a}^{3}}{(6.67\ \times \ {10}^{-11}\ \frac{{\text{m}}^{3}}{\text{kg}\cdot \ {\text{sec}}^{2}})(7.35\ x\ {10}^{22}\text{kg}+5.97\ x\ {10}^{24}\text{kg})} \\ 5.563\ \times \ {10}^{12} & = & \frac{4{\pi }^{2}{a}^{3}}{(6.67\ \times \ {10}^{-11}\ {\text{m}}^{3})(6.04\ x\ {10}^{24})} \\ (5.563\ \times \ {10}^{12})(6.67\ \times \ {10}^{-11}\ {\text{m}}^{3})(6.04\ \times \ {10}^{24}) & = & 4{\pi }^{2}{a}^{3} \\ {a}^{3} & = & \frac{2.241\ \times \ {10}^{27}}{4{\pi }^{2}}{\text{m}}^{3} \\ a & = & 3.84\ \times \ {10}^{8}\text{m} \\ & \approx & 384,000\ \text{km.}\end{array}\]

8. Titan is the largest moon of Saturn. The mass of Titan is approximately \(1.35\ \times \ {10}^{23}\) kg. The mass of Saturn is approximately \(5.68\ \times \ {10}^{26}\) kg. Titan takes approximately 16 days to orbit Saturn. Use this information, along with the universal gravitation constant \(G=6.67\ \times \ {10}^{-11}{\text{m}}^{3}\text{/}\text{kg}\cdot {\text{sec}}^{2}\) to estimate the distance from Titan to Saturn.

   Rivela la risposta

   \(a=1.224\ \times \ {10}^{9}\text{m}\approx 1,224,000\ \text{km}\)

9. We now return to the chapter opener, which discusses the motion of Halley’s comet around the Sun. Kepler’s first law states that Halley’s comet follows an elliptical path around the Sun, with the Sun as one focus of the ellipse. The period of Halley’s comet is approximately 76.1 years, depending on how closely it passes by Jupiter and Saturn as it passes through the outer solar system. Let’s use \(T=76.1\) years. What is the average distance of Halley’s comet from the Sun?

   Rivela la risposta

   Using the equation \({T}^{2}={D}^{3}\) with \(T=76.1,\) we obtain \({D}^{3}=5791.21,\) so \(D\approx 17.96\) A.U. This comes out to approximately \(1.67\ \times \ {10}^{9}\) mi.

   A natural question to ask is: What are the maximum (aphelion) and minimum (perihelion) distances from Halley’s Comet to the Sun? The eccentricity of the orbit of Halley’s Comet is 0.967 (Source: http://nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html). Recall that the formula for the eccentricity of an ellipse is \(e=c\text{/}a,\) where a is the length of the semimajor axis and c is the distance from the center to either focus. Therefore, \(0.967=c\text{/}17.96\) and \(c\approx 17.37\) A.U. Subtracting this from a gives the perihelion distance \(p=a-c=17.96-17.37=0.59\) A.U. According to the National Space Science Data Center (Source: http://nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html), the perihelion distance for Halley’s comet is 0.587 A.U. To calculate the aphelion distance, we add

   \[P=a+c=17.96+17.37=35.33\ \text{A}\text{.U}\text{.}\]

   This is approximately \(3.3\ \times \ {10}^{9}\) mi. The average distance from Pluto to the Sun is 39.5 A.U. (Source: http://www.oarval.org/furthest.htm), so it would appear that Halley’s Comet stays just within the orbit of Pluto.

10. Given \(\text{r}(t)=(3{t}^{2}-2)\text{i}+(2t-\text{sin}(t))\text{j},\) find the velocity of a particle moving along this curve.

    Rivela la risposta

    \(\text{v}(t)=(6t)\text{i}+(2-\text{cos}(t))\text{j}\)

11. Given \(\text{r}(t)=(3{t}^{2}-2)\text{i}+(2t-\text{sin}(t))\text{j},\) find the acceleration vector of a particle moving along the curve in the preceding exercise.

12. \(\text{r}(t)=〈3\ \text{cos}\ t,3\ \text{sin}\ t,{t}^{2}〉\)

    Rivela la risposta

    \(\text{v}(t)=〈-3\ \text{sin}\ t,3\ \text{cos}\ t,2t〉,\) \(\text{a}(t)=〈-3\ \text{cos}\ t,-3\ \text{sin}\ t,2〉,\) \(\text{speed}=\sqrt{9+4{t}^{2}}\)

13. \(\text{r}(t)={e}^{\text{-}t}\text{i}+{t}^{2}\text{j}+\text{tan}\ t\ \text{k}\)

14. \(\text{r}(t)=2\ \text{cos}\ t\ \text{j}+3\ \text{sin}\ t\ \text{k}.\)

    Rivela la risposta

    \(v(t)=-2\sin \ t\ j+3\ \cos \ t\ k,\) \(\text{a}(t)=-2\ \text{cos}\ t\ \text{j}-3\ \text{sin}\ t\ \text{k},\) \(\text{speed}=\sqrt{4\ {\text{sin}}^{2}t+9{\text{cos}}^{2}t}\)

15. \(\text{r}(t)=〈{t}^{2}-1,t〉\)

16. \(\text{r}(t)=〈{e}^{t},{e}^{\text{-}t}〉\). The graph is shown here:

    Rivela la risposta

    \(\text{v}(t)={e}^{t}\text{i}-{e}^{\text{-}t}\text{j},\) \(\text{a}(t)={e}^{t}\text{i}+{e}^{\text{-}t}\text{j},\) speed = \(‖\text{v}(t)‖=\sqrt{{e}^{2t}+{e}^{-2t}}\)

17. \(\text{r}(t)=〈\text{sin}\ t,t,\text{cos}\ t〉.\)

18. The position function of an object is given by \(\text{r}(t)=〈{t}^{2},5t,{t}^{2}-16t〉.\) At what time is the speed a minimum?

    Rivela la risposta

    \(t=4\)

19. Let \(\text{r}(t)=r\ \text{cosh}(\omega t)\text{i}+r\ \text{sinh}(\omega t)\text{j}.\) Find the velocity and acceleration vectors and show that the acceleration is proportional to \(\text{r}(t).\)

20. Find the equations for the velocity, acceleration, and speed of the particle at any time.

    Rivela la risposta

    \(\text{v}(t)=(\omega -\omega \ \text{cos}(\omega t))\ \text{i}+(\omega \ \text{sin}(\omega t))\ \text{j},\)
    \(\text{a}(t)=({\omega }^{2}\text{sin}(\omega t))\ \text{i}+({\omega }^{2}\text{cos}(\omega t))\ \text{j},\)
    \(\text{speed}=\sqrt{{\omega }^{2}-2{\omega }^{2}\text{cos}(\omega t)+{\omega }^{2}{\text{cos}}^{2}(\omega t)+{\omega }^{2}{\text{sin}}^{2}(\omega t)}\ \text{=}\ \sqrt{2{\omega }^{2}(1-\text{cos}(\omega t))}\)

21. The velocity and acceleration vectors

22. The glider’s speed at any time

    Rivela la risposta

    \(‖\text{v}(t)‖=\sqrt{9+4{t}^{2}}\)

23. The times, if any, at which the glider’s acceleration is orthogonal to its velocity

24. The velocity of the particle

    Rivela la risposta

    \(\text{v}(t)=〈{e}^{-5t}(\text{cos}\ t-5\ \text{sin}\ t),\text{-}{e}^{-5t}(\text{sin}\ t+5\ \text{cos}\ t),-20{e}^{-5t}〉\)

25. The speed of the particle

26. The acceleration of the particle

    Rivela la risposta

    \(\text{a}(t)={〈e}^{-5t}(\text{-}\text{sin}\ t-5\ \text{cos}\ t)-5{e}^{-5t}(\text{cos}\ t-5\ \text{sin}\ t),\) \(\text{-}{e}^{-5t}(\text{cos}\ t-5\ \text{sin}\ t)+5{e}^{-5t}(\text{sin}\ t+5\ \text{cos}\ t),100{e}^{-5t}〉\)

27. Find the maximum speed of a point on the circumference of an automobile tire of radius 1 ft when the automobile is traveling at 55 mph.

28. At what time does the projectile reach maximum height?

    Rivela la risposta

    44.185 sec

29. What is the approximate maximum height of the projectile?

30. At what time is the maximum range of the projectile attained?

    Rivela la risposta

    \(t=88.37\) sec

31. What is the maximum range?

32. What is the total flight time of the projectile?

    Rivela la risposta

    88.37 sec

33. Determine the maximum height of the projectile.

34. Determine the range of the projectile.

    Rivela la risposta

    The range is approximately 886.29 m.

35. A golf ball is hit in a horizontal direction off the top edge of a building that is 100 ft tall. How fast must the ball be launched to land 450 ft away?

36. A projectile is fired from ground level at an angle of 8° with the horizontal. The projectile is to have a range of 50 m. Find the minimum velocity necessary to achieve this range.

    Rivela la risposta

    \(\text{v}=42.16\) m/sec

37. Prove that an object moving in a straight line at a constant speed has an acceleration of zero.

38. The acceleration of an object is given by \(\text{a}(t)=t\ \text{j}+t\ \text{k}.\) The velocity at \(t=1\) sec is \(\text{v}(1)=5\text{j}\) and the position of the object at \(t=1\) sec is \(\text{r}(1)=0\text{i}+0\text{j}+0\text{k}.\) Find the object’s position at any time.

    Rivela la risposta

    \(\text{r}(t)=0\text{i}+(\frac{1}{6}{t}^{3}+4.5t-\frac{14}{3})\ \text{j}+(\frac{{t}^{3}}{6}-\frac{1}{2}t+\frac{1}{3})\ \text{k}\)

39. Find \(\text{r}(t)\) given that \(\text{a}(t)=-32\text{j},\) \(\text{v}(0)=600\sqrt{3}\text{i}+600\text{j},\) and \(\text{r}(0)=0.\)

40. Find the tangential and normal components of acceleration for \(\text{r}(t)=a\ \text{cos}(\omega t)\text{i}+a\ \text{sin}(\omega t)\text{j}\) at \(t=0.\)

    Rivela la risposta

    \({a}_{T}=0,\) \({a}_{N}=a{\omega }^{2}\)

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Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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