# Planar Motion

### Position and Trajectory

Suppose a particle moves in a plane. Then its position can be expressed as

${\mathbf{r} = x\mathbf{i} + y\mathbf{j}},$

where $$x,y$$ are the Cartesian coordinates and $$\mathbf{i},\mathbf{j}$$ are the unit vectors along the $$x-$$ and $$y-$$ coordinate axes, respectively.

As the coordinates $$x,y$$ depend on time, we can write the position vector in parametric form:

${\mathbf{r} = x\left( t \right)\mathbf{i} + y\left( t \right)\mathbf{j}}.$

This vector equation defines the trajectory of the particle.

### Velocity and Speed

The velocity $$\mathbf{v}$$ for a particle is defined as the time derivative of the particle’s position vector, that is

${\mathbf{v} = \frac{{d\mathbf{r}}}{{dt}}}.$

In coordinate form, it is given by

${\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j}},$

where $${v_x} = \large{\frac{{dx}}{{dt}}}\normalsize,$$ $${v_y} = \large{\frac{{dy}}{{dt}}}\normalsize$$ are components of the velocity vector.

The particle’s speed $$v$$ is a scalar quantity. It is defined as the magnitude of the velocity:

${{v = \sqrt {v_x^2 + v_y^2} }={ \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} .}}$

### Acceleration

The acceleration $$\mathbf{a}$$ for a particle is defined as the time derivative of the particle’s velocity vector:

${\mathbf{a} = \frac{{d\mathbf{v}}}{{dt}} }= \frac{{{d^2}\mathbf{r}}}{{d{t^2}}}.$

In case when $$\mathbf{v}$$ is written in coordinate form, the acceleration is given by

${\mathbf{a} = \frac{{d{v_x}}}{{dt}}\mathbf{i} + \frac{{d{v_y}}}{{dt}}\mathbf{j} }={ \frac{{{d^2}x}}{{d{t^2}}}\mathbf{i} + \frac{{{d^2}y}}{{d{t^2}}}\mathbf{j},}$

where $${a_x} = \large{\frac{{d{v_x}}}{{dt}}}\normalsize = \large{\frac{{{d^2}x}}{{d{t^2}}}}\normalsize,$$ $${a_y} = \large{\frac{{d{v_y}}}{{dt}}}\normalsize = \large{\frac{{{d^2}y}}{{d{t^2}}}}\normalsize$$ are components of the acceleration vector.

The magnitude of the acceleration is computed as

${a = \sqrt {a_x^2 + a_y^2} }={ \sqrt {{{\left( {\frac{{d{v_x}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{v_y}}}{{dt}}} \right)}^2}} }={ \sqrt {{{\left( {\frac{{{d^2}x}}{{d{t^2}}}} \right)}^2} + {{\left( {\frac{{{d^2}y}}{{d{t^2}}}} \right)}^2}} .}$

### Motion with Constant Acceleration

Suppose a particle or object moves with a constant acceleration $$\mathbf{a}:$$

${\mathbf{a} = \frac{{d\mathbf{v}}}{{dt}} = \text{const}.}$

Then the velocity at time $$t$$ is written as

${\mathbf{v}\left( t \right) = {\mathbf{v}_0} + \mathbf{a}t,}$

where $${\mathbf{v}_0}$$ is the initial velocity vector at $$t = 0.$$

The position at time $$t$$ is described by the equation

${\mathbf{r}\left( t \right) = {\mathbf{r}_0} + {\mathbf{v}_0}t + \frac{1}{2}\mathbf{a}{t^2}},$

where $${\mathbf{r}_0}$$ is the initial position at $$t = 0.$$

### Free Fall Motion

Free fall is motion of a body with a constant acceleration caused by gravity. Such an object moves with the downward acceleration

${\mathbf{a} = – g\mathbf{j}},$

where $$g = 9.8\,\large{\frac{\text{m}}{\text{s}^2}}\normalsize$$ is the acceleration due to gravity, $$\mathbf{j}$$ is a unit vector pointing vertically upward.

The vertical velocity $$v\left( t \right)$$ and the position (or altitude) $$y\left( t \right)$$ during a free fall are given by the equations

${v\left( t \right) = {v_0} – gt\;\;\left({\frac{\text{m}}{\text{s}}}\right)},$

${y\left( t \right) = {y_0} + {v_0}t – \frac{1}{2}g{t^2}\;\;\left({\text{m}}\right)},$

where $${v_0}$$ is the initial velocity.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A particle is moving in $$xy-$$plane according to the equations $$x = t,$$ $$y = {t^3},$$ where $$x,y$$ are measured in meters. Find the particle’s velocity and speed at $$t = 1\,\text{s}.$$

### Example 2

Assume that an object moves on a trajectory according to the equations $$x\left( t \right) = t + \cos t,$$ $$y\left( t \right) = t – \sin t.$$ Find the magnitude of the acceleration vector.

### Example 3

A particle moves along the hyperbola $$y = \large{\frac{{12}}{x}}\normalsize,$$ so that the $$x-$$coordinate is increasing at a constant rate of $$3$$ meters per second. Find the speed of the particle when it is at the point $$\left( {3,4} \right).$$

### Example 4

A particle is moving along the curve given by the parametric equations $$x = 1 + t,$$ $$y = 1 – t.$$ Determine the $$xy-$$equation of the trajectory and speed along it.

### Example 5

A particle moves on a trajectory described by the parametric equations $$x\left( t \right) = {2^t},$$ $$y\left( t \right) = {8^t},$$ $$t \ge 0.$$ What is the shape of the trajectory?

### Example 6

A particle’s position is given by the equations $$x\left( t \right) = \large{\frac{1}{2}}\normalsize – {t^2},$$ $$y\left( t \right) = \sqrt 2 t,$$ where $$x,y$$ are measured in meters. Determine the particle’s distance $$d$$ from the origin as a function of time $$t.$$

### 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 $$t = \large{\frac{\pi }{6}}\normalsize.$$

### 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 $$t = 2\,\text{s}.$$

### Example 9

A ball is thrown up with a speed of $${v_0} = 19.6\,\large{\frac{\text{m}}{\text{s}}}\normalsize.$$
1. What is the time it takes to hit the ground? The acceleration of gravity is $$9.8\,\large{\frac{\text{m}}{\text{s}^2}}\normalsize;$$
2. What is the ball’s maximum height above the ground?

### Example 10

Two balls are thrown horizontally from the same point in the opposite direction with the initial speeds $${v_{10}} = 4\,\large{\frac{\text{m}}{\text{s}}}\normalsize,$$ $${v_{20}} = 9\,\large{\frac{\text{m}}{\text{s}}}\normalsize$$ (Figure $$4$$). Find the distance $$d$$ between the balls when their velocity vectors are perpendicular to each other.

### 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.

### Example 1.

A particle is moving in $$xy-$$plane according to the equations $$x = t,$$ $$y = {t^3},$$ where $$x,y$$ are measured in meters. Find the particle’s velocity and speed at $$t = 1\,\text{s}.$$

Solution.

From the definition of velocity,

${\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j}},$

where

${\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( t \right) = 1,\;\;}\kern0pt{\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^3}} \right) = 3{t^2}.}$

Hence, the velocity vector at $$t = 1$$ is given by

${\mathbf{v}\left( {t = 1} \right) = \mathbf{i} + 3\mathbf{j}.}$

The speed at this time is equal to

${{\left| {\mathbf{v}} \right| = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} }}={ \sqrt {{1^2} + {3^2}} }={ \sqrt {10}\,\frac{\text{m}}{\text{s}}.}$

### Example 2.

Assume that an object moves on a trajectory according to the equations $$x\left( t \right) = t + \cos t,$$ $$y\left( t \right) = t – \sin t.$$ Find the magnitude of the acceleration vector.

Solution.

We differentiate $$x\left( t \right)$$ and $$y\left( t \right)$$ twice to find the acceleration of the object:

${x^\prime\left( t \right) = \left( {t + \cos t} \right)^\prime }={ 1 – \sin t,}$

${x^{\prime\prime}\left( t \right) = \left( {1 – \sin t} \right)^\prime }={ – \cos t,}$

${y^\prime\left( t \right) = \left( {t – \sin t} \right)^\prime }={ 1 + \cos t,}$

${y^{\prime\prime}\left( t \right) = \left( {1 + \cos t} \right)^\prime }={ – \sin t.}$

Now we can compute the magnitude of the acceleration vector:

${a = \left| \mathbf{a} \right| }={ \sqrt {{{\left( {x^{\prime\prime}\left( t \right)} \right)}^2} + {{\left( {y^{\prime\prime}\left( t \right)} \right)}^2}} }={ \sqrt {{{\left( { – \cos t} \right)}^2} + {{\left( { – \sin t} \right)}^2}} }={ \sqrt {{{\cos }^2}t + {{\sin }^2}t} }={ 1.}$

### Example 3.

A particle moves along the hyperbola $$y = \large{\frac{{12}}{x}}\normalsize,$$ so that the $$x-$$coordinate is increasing at a constant rate of $$3$$ meters per second. Find the speed of the particle when it is at the point $$\left( {3,4} \right).$$

Solution.

First, we determine the velocity vector of the particle using the formula

$\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j}.$

The derivative $$\large{\frac{{dx}}{{dt}}}\normalsize$$ is known:

$\frac{{dx}}{{dt}} = 3\,\frac{\text{m}}{\text{s}}.$

Compute the derivative $$\large{\frac{{dy}}{{dt}}}\normalsize$$ by the chain rule:

${\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{12}}{x}} \right) }={ – \frac{{12}}{{{x^2}}}\frac{{dx}}{{dt}}.}$

At the given point we have

$\frac{{dy}}{{dt}} = – \frac{{12}}{{{3^2}}} \cdot 3 = – 4\,\frac{\text{m}}{\text{s}}.$

Now we can find the particle’s speed:

${v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} }={ \sqrt {{3^2} + {{\left( { – 4} \right)}^2}} }={ 5\,\frac{\text{m}}{\text{s}}.}$

### Example 4.

A particle is moving along the curve given by the parametric equations $$x = 1 + t,$$ $$y = 1 – t.$$ Determine the $$xy-$$equation of the trajectory and speed along it.

Solution.

We solve the first equation for $$t$$ and plug it into the second equation:

${x = 1 + t,\;\;} \Rightarrow {t = x – 1,\;\;} \Rightarrow {y = 1 – \left( {x – 1} \right),\;\;} \Rightarrow {y = 2 – x.}$

Hence, the particle’s trajectory is the straight line $$y = 2 – x.$$ Find the velocity vector:

$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {1 + t} \right) = 1,$

$\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {1 – t} \right) = – 1.$

Then

${\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j} }={ 1 \cdot \mathbf{i} + \left( { – 1} \right) \cdot \mathbf{j} }={ \mathbf{i} – \mathbf{j}.}$

The speed is equal to

${v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} }={ \sqrt {{1^2} + {{\left( { – 1} \right)}^2}} }={ \sqrt 2 .}$

$y = 2 – x,\;v = \sqrt 2 .$

### Example 5.

A particle moves on a trajectory described by the parametric equations $$x\left( t \right) = {2^t},$$ $$y\left( t \right) = {8^t},$$ $$t \ge 0.$$ What is the shape of the trajectory?

Solution.

From the first equation, we find:

${x\left( t \right) = {2^t},\;\;} \Rightarrow {t = {\log _2}x.}$

Substitute this into the second equation to obtain the $$xy-$$equation of the curve:

${y = {8^t},\;\;} \Rightarrow {y = {8^{{{\log }_2}x}} }={ {\left( {{2^3}} \right)^{{{\log }_2}x}} }={ {2^{{{\log }_2}{x^3}}} }={ {x^3}.}$

As $$t \ge 0,$$ then $$x \ge {2^0} = 1.$$

Hence, the given curve is the cubic parabola $$y = {x^3}$$ where $$x \ge 1.$$

### Example 6.

A particle’s position is given by the equations $$x\left( t \right) = \large{\frac{1}{2}}\normalsize – {t^2},$$ $$y\left( t \right) = \sqrt 2 t,$$ where $$x,y$$ are measured in meters. Determine the particle’s distance $$d$$ from the origin as a function of time $$t.$$

Solution.

The distance from the origin is defined as

$d\left(t\right) = \sqrt {{x^2}\left( t \right) + {y^2}\left( t \right)} .$

Substituting $$x\left( t \right),$$ $$y\left( t \right),$$ we obtain:

${d\left(t\right) = \sqrt {{{\left( {\frac{1}{2} – {t^2}} \right)}^2} + {{\left( {\sqrt 2 t} \right)}^2}} }={ \sqrt {\frac{1}{4} – {t^2} + {t^4} + 2{t^2}} }={ \sqrt {{t^4} + {t^2} + \frac{1}{4}} }={ \sqrt {{{\left( {{t^2} + \frac{1}{2}} \right)}^2}} }={ {{t^2} + \frac{1}{2}}\,\left(\text{m}\right).}$

### 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 $$t = \large{\frac{\pi }{6}}\normalsize.$$

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 $$t = \large{\frac{\pi }{6}}\normalsize,$$ 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:

${\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {3{t^3} – 3{t^2}} \right) }={ 9{t^2} – 6t,}$

${\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {20{t^2} + 2t} \right) }={ 40t + 2.}$

The 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 $${v_0} = 19.6\,\large{\frac{\text{m}}{\text{s}}}\normalsize.$$
1. What is the time it takes to hit the ground? The acceleration of gravity is $$9.8\,\large{\frac{\text{m}}{\text{s}^2}}\normalsize;$$
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 $$t:$$

${t\left( {{v_0} – \frac{{gt}}{2}} \right) = 0,\;\;} \Rightarrow {{t_1} = 0,\;}\kern0pt{{t_2} = \frac{{2{v_0}}}{g}.}$

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 $${v_{10}} = 4\,\large{\frac{\text{m}}{\text{s}}}\normalsize,$$ $${v_{20}} = 9\,\large{\frac{\text{m}}{\text{s}}}\normalsize$$ (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:

${{v_{1x}} = – {v_{10}},\;}\kern0pt{{v_{1y}} = – gt,}$

${{v_{2x}} = {v_{20}},\;}\kern0pt{{v_{2y}} = – gt.}$

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 $$t = \large{\frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}}\normalsize$$ 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

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