# 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 $${v_0} = 19.6\,\frac{\text{m}}{\text{s}}.$$

1. What is the time it takes to hit the ground? The acceleration of gravity is $$9.8\,\frac{\text{m}}{\text{s}^2};$$
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.

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 $$a \gt 0,$$ $$b \gt 0.$$

1. Determine the particle's trajectory $$y\left( x \right)$$ and sketch its graph.
2. Determine the speed of the particle as a function of time.
3. 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

$y = \frac{b}{{{a^2}}}{x^2}.$

$$2.$$ 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

$\tan \varphi = \frac{{2bt}}{a}.$

Then

$\varphi = \arctan \left( {\frac{{2bt}}{a}} \right).$