Planar Motion
Solved Problems
Example 7.
A particle is moving along the curve given by the parametric equations \[x = \tan t, y = \sec t.\] Find the particle's speed at \(t = \frac{\pi }{6}.\)
Solution.
We take the derivatives of the coordinates \(x\) and \(y\) with respect to time \(t:\)
\[\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\tan t} \right) = \frac{1}{{{{\cos }^2}t}},\]
\[\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\sec t} \right) = \frac{d}{{dt}}\left( {\frac{1}{{\cos t}}} \right) = - \frac{1}{{{{\cos }^2}t}} \cdot \left( { - \sin t} \right) = \frac{{\sin t}}{{{{\cos }^2}t}}.\]
Calculate the particle's speed by the formula
\[v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} .\]
This yields:
\[v = \sqrt {{{\left( {\frac{1}{{{{\cos }^2}t}}} \right)}^2} + {{\left( {\frac{{\sin t}}{{{{\cos }^2}t}}} \right)}^2}} = \frac{{\sqrt {1 + {{\sin }^2}t} }}{{{{\cos }^2}t}}.\]
Substituting the time value \(t = \frac{\pi }{6},\) we get
\[v = \frac{{\sqrt {1 + {{\sin }^2}\frac{\pi }{6}} }}{{{{\cos }^2}\frac{\pi }{6}}} = \frac{{\sqrt {1 + {{\left( {\frac{1}{2}} \right)}^2}} }}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = \frac{{\sqrt {1 + \frac{1}{4}} }}{{\frac{3}{4}}} = \frac{{\frac{{\sqrt 5 }}{2}}}{{\frac{3}{4}}} = \frac{{2\sqrt 5 }}{3}.\]
Example 8.
A particle moves in the \(xy-\)plane, so that its coordinates at time \(t \ge 0\) are
\[x\left( t \right) = 3{t^3} - 3{t^2}, y\left( t \right) = 20{t^2} + 2t,\]
where \(x,y\) are in meters, \(t\) in seconds. Find the magnitude of the particle's acceleration at \(t = 2\,\text{s}.\)
Solution.
First, we compute the particle's velocity:
\[\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {3{t^3} - 3{t^2}} \right) = 9{t^2} - 6t,\]
\[\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {20{t^2} + 2t} \right) = 40t + 2.\]
Then the velocity vector is given by
\[\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j} = \left( {9{t^2} - 6t} \right)\mathbf{i} + \left( {40t + 2} \right)\mathbf{j}.\]
Differentiating once more, we get the acceleration:
\[\frac{{{d^2}x}}{{d{t^2}}} = \frac{d}{{dt}}\left( {9{t^2} - 6t} \right) = 18t - 6,\]
\[\frac{{{d^2}y}}{{d{t^2}}} = \frac{d}{{dt}}\left( {40t + 2} \right) = 40.\]
Hence
\[\mathbf{a} = \frac{{{d^2}x}}{{d{t^2}}}\mathbf{i} + \frac{{{d^2}y}}{{d{t^2}}}\mathbf{j} = \left( {18t - 6} \right)\mathbf{i} + 40\mathbf{j}.\]
Substitute \(t = 2\,\text{s}:\)
\[\mathbf{a} = 30\mathbf{i} + 40\mathbf{j}.\]
Then the magnitude of the acceleration vector is equal to
\[a = \left| \mathbf{a} \right| = \sqrt {{{30}^2} + {{40}^2}} = 50\,\frac{\text{m}}{{{\text{s}^2}}}.\]
Example 9.
A ball is thrown up with a speed of \({v_0} = 19.6\,\frac{\text{m}}{\text{s}}.\)
What is the time it takes to hit the ground? The acceleration of gravity is \(9.8\,\frac{\text{m}}{\text{s}^2};\)
What is the ball's maximum height above the ground?
Solution.
\(1.\) The \(y-\)coordinate of the ball is defined by the equation
\[y\left( t \right) = {v_0}t - \frac{{g{t^2}}}{2}.\]
When the ball hits the ground, the \(y-\)coordinate is equal to zero. Therefore, we can write
\[y\left( t \right) = {v_0}t - \frac{{g{t^2}}}{2} = 0.\]
Solve this equation for \(t:\)
\[t\left( {{v_0} - \frac{{gt}}{2}} \right) = 0,\;\; \Rightarrow {t_1} = 0,\;{t_2} = \frac{{2{v_0}}}{g}.\]
The second root gives the time when the ball hits the ground:
\[{t_2} = \frac{{2{v_0}}}{g} = \frac{{2 \cdot 19.6}}{{9.8}} = 4\,\text{s}.\]
\(2.\) The time taken to reach the maximum point can be found from the equation \(v\left( t \right) = 0.\) Hence,
\[v\left( t \right) = {v_0} - gt = 0,\;\; \Rightarrow t = \frac{{{v_0}}}{g}.\]
Calculate the maximum height:
\[{y_{\max }} = {v_0}t - \frac{{g{t^2}}}{2} = {v_0}\left( {\frac{{{v_0}}}{g}} \right) - \frac{g}{2}{\left( {\frac{{{v_0}}}{g}} \right)^2} = \frac{{v_0^2}}{g} - \frac{{v_0^2}}{{2g}} = \frac{{v_0^2}}{{2g}} = \frac{{{{\left( {19.6} \right)}^2}}}{{2 \cdot 9.8}} = 19.6\,\text{m}\]
Example 10.
Two balls are thrown horizontally from the same point in the opposite direction with the initial speeds \({v_{10}} = 4\,\frac{\text{m}}{\text{s}},\) \({v_{20}} = 9\,\frac{\text{m}}{\text{s}}\) (Figure \(4\)). Find the distance \(d\) between the balls when their velocity vectors are perpendicular to each other.
Solution.
Figure 4. First, we write the velocity components for each ball:
\[{v_{1x}} = - {v_{10}},\; {v_{1y}} = - gt,\]
\[{v_{2x}} = {v_{20}},\; {v_{2y}} = - gt.\]
Hence, the velocity vectors are given by
\[\mathbf{v}_1 = - {v_{10}}\mathbf{i} - gt\mathbf{j},\]
\[\mathbf{v}_2 = {v_{20}}\mathbf{i} - gt\mathbf{j}.\]
When the velocity vectors are perpendicular, their dot product is equal to zero, so that
\[{\mathbf{v}_1} \bot {\mathbf{v}_2},\;\; \Rightarrow {\mathbf{v}_1} \cdot {\mathbf{v}_2} = 0,\;\; \Rightarrow \left( { - {v_{10}}} \right){v_{20}} + \left( { - gt} \right)\left( { - gt} \right) = 0,\;\; \Rightarrow {g^2}{t^2} = {v_{10}}{v_{20}},\;\;\Rightarrow t = \frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}.\]
Note that both the balls are always on the same height as their vertical velocity components are equal. Therefore, the line \(d\) is horizontal.
The distance between the balls at the time instant \(t = \frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}\) is equal to
\[d = {v_{10}}t + {v_{20}}t = \left( {{v_{10}} + {v_{20}}} \right)t = \left( {{v_{10}} + {v_{20}}} \right)\frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}.\]
Substituting the known values, we obtain:
\[d = \left( {4 + 9} \right)\frac{{\sqrt {4 \cdot 9} }}{{9.8}} = 7.96\,\text{m}\]
Example 11.
A particle moves in the \(xy-\) plane according to the law \[x = at, y = b{t^2},\] where \(a \gt 0,\) \(b \gt 0.\)
Determine the particle's trajectory \(y\left( x \right)\) and sketch its graph.
Determine the speed of the particle as a function of time.
Find the angle \(\phi\) between the velocity vector and the \(x-\)axis.
Solution.
\(1.\) Let's solve the first equation for \(t\) and substitute it into the second equation:
\[x = at,\;\; \Rightarrow t = \frac{x}{a},\;\; \Rightarrow y = b{\left( {\frac{x}{a}} \right)^2} = \frac{b}{{{a^2}}}{x^2}.\]
So the particle's trajectory is the right branch of the parabola
\[y = \frac{b}{{{a^2}}}{x^2}.\]
Figure 5. \(2.\) To determine the speed of the particle, we differentiate the coordinates \(x,y\) with respect to time \(t:\)
\[\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {at} \right) = a,\]
\[\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {b{t^2}} \right) = 2bt.\]
Therefore, the particle's velocity vector is given by
\[\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j} = a\mathbf{i} + 2bt\mathbf{j}.\]
Since the speed is the absolute value of the velocity, we have
\[v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} = \sqrt {{a^2} + 4{b^2}{t^2}} .\]
\(3.\) The velocity vector is written in the form
\[\mathbf{v} = a\mathbf{i} + 2bt\mathbf{j}.\]
Therefore, the tangent of the angle \(\varphi\) is given by
\[\tan \varphi = \frac{{2bt}}{a}.\]
Figure 6. Then
\[\varphi = \arctan \left( {\frac{{2bt}}{a}} \right).\]
