Movement of a Particle I

Task number: 1988

The position vector of a particle changes with time according to relation:

\[\vec{r}\left(t\right)\,=\,15\,\mathrm{m \cdot s^{-2}}\cdot t^2\vec{i}+(4\,\mathrm{m}-20\,\mathrm{m \cdot s^{-2}}\cdot t^2)\vec{j}\]

where \(\vec{i}\), \(\vec{j}\)are unit vectors in directions of the coordinate axes x and y.

Examine the movement of the particle according to the following points:

a) Which relations describe the movement of the particle in directions of the coordinate axes x, y, z?

b) At what time does the particle cross the y axis?

c) At what time does the particle cross the x axis?

d) How does the velocity of the particle and its components change with time?

e) What motions does the particle perform in directions of the axes x, y, z?

f) How does the speed of the particle change with time? (speed refers to the magnitude of velocity)

g) Is the movement of the particle uniform?

h) How does the acceleration of the particle and its components change with time?

i) How does the magnitude of acceleration of the particle change with time?

j) What is the speed and the acceleration at the moment the particle crosses:

1) x 2) y?

Note: Parametric equations should be written in the following form:

\[y\,=\,1\,\mathrm{m} - 2\,\mathrm{m \cdot s^{-1}}\cdot t\]

We do not use units in relations to simplify the notation.

Given values

\(\vec{i}\), \(\vec{j}\), \(\vec{k}\)	unit vectors in directions of the coordinate axes x, y, z
\[\vec{r}\left(t\right)\,=\,15\ t^2\vec{i}+(4-20t^2)\vec{j}\]	position vector of the particle at time t
t₁ = ? (s)	time at which the particle crosses the y axis
t₂ = ? (s)	time at which the particle crosses the x axis
v = ? (m·s^-1)	velocity of the particle at the time it crosses the x axis (resp. the y axis)
a = ? (m·s^-2)	acceleration of the particle at the time crosses the x axis (resp. the y axis)

Hint 1 for a): Movement of the particle in directions of the coordinate axes
You can easily obtain the individual relations for movement in directions of the coordinate axes from the given position vector.
Solution of Hint 1
We can write the position vector as:
\[\vec{r}\left(t\right)\,=\, x\vec{i}+ y\vec{j}+ z\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.

From here:
\[x\left(t\right)\,=\,15t^{2}\,,\] \[y\left(t\right)\,=\,4-20t^{2}\,,\] \[z\left(t\right)\,=\,0\,.\]
Hint 2 for b): Time at which the particle crosses the y axis
How can you determine the time t₁ at which the particle crosses the y axis?

What is the x coordinate at that moment?
Solution of Hint 2
The particle will cross the y axis at the moment its x coordinate is equal to 0:
\[x\left(t\right)\,=\,15t^{2}\,,\] \[0\,=\,15t_1^{2}\,,\] \[t_1\,=\,0\,\mathrm{s}\,.\]
Hint 3 for c): Time at which the particle crosses the x x axis
How can you determine time t₂ at which the particle crosses the x axis?

What is the y coordinate at that moment?
Solution of Hint 3
The particle crosses the x axis at the moment its y coordinate is equal to 0:
\[y\left(t\right)\,=\,4-20t^{2}\,,\] \[0\,=\,4-20t_2^{2}\,,\] \[t_2^{2}\,=\,\frac{1}{5}\,,\] \[t_2\,=\,\frac{1}{\sqrt{5}}\,\mathrm{s}\,.\]
(The mathematical solution \(t_2\,=\,\frac{-1}{\sqrt{5}}\,\mathrm{s}\) bears no physical significance.)
Hint 4 for d): Velocity of the particle
You know the corresponding relations describing the movement of the particle in directions of the coordinate axes.

How can you use them and the definition of velocity to obtain the behaviour of the velocity of the particle?
Solution of Hint 4
From the definition of velocity we obtain:

\[\vec{v}(t)\,=\,\frac{\mathrm{d}\vec{r}\left(t\right)}{\mathrm{d}t}\,=\,\frac{\mathrm{d}x\left(t\right)}{\mathrm{d}t}\vec{i}+\frac{\mathrm{d}y\left(t\right)}{\mathrm{d}t}\vec{j}+\frac{\mathrm{d}z\left(t\right)}{\mathrm{d}t}\vec{k}\,,\] \[v_\mathrm{x}\,=\,\frac{\mathrm{d}x}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}\left(15t^{2}\right)\,=\, 30t\,,\] \[v_\mathrm{y}\,=\,\frac{\mathrm{d}y}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}\left(4-20t^{2}\right)\,=\, -40t \,,\] \[v_\mathrm{z}\,=\,\frac{\mathrm{d}z}{\mathrm{d}t}\,=\,\frac{\mathrm{d}}{\mathrm{d}t}\left(0\right)\,=\, 0\,,\] \[\vec{v}\left(t\right)\,=\, 30t\vec{i}+\left(-40t\right)\vec{j}+0\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.
Hint 5 for e): Movement of the particle in directions of the coordinate axes
What motions does the particle perform in directions of the axes x, y, z?

Look at the relations with which the movement of the particle in directions of the coordinate axes is described and how the coordinates of velocity change with time.
Solution of Hint 5
In the direction of x:
\[x\left(t\right)\,=\,15t^{2}\,,\] \[v_x\left(t\right)\,=\, 30t\,.\]
The movement is uniformly accelerated.

In the direction of y:
\[y\left(t\right)\,=\,4-20t^{2}\,,\] \[v_y\left(t\right)\,=\, -40t\,.\]
The movement is uniformly accelerated in the opposite direction.

In the direction of z:
\[z\left(t\right)\,=\, 0\,,\] \[v_z\left(t\right)\,=\, 0\,.\]
The particle is at rest.
Hint 6 for f): Speed of the particle
You know the behaviour of the velocity of the particle and its coordinates. How can you determine the behaviour of its magnitude?
Solution of Hint 6
From the definition of speed we obtain:

\[v\left(t\right)\,=\,\left|\vec{v}\left(t\right)\right|\,=\,\sqrt{v_\mathrm{x}^{2}\left(t\right)+v_\mathrm{y}^{2}\left(t\right)+v_\mathrm{z}^{2}\left(t\right)}\,,\] \[v\left(t\right)\,=\,\sqrt{(900t^{2}+1600t^{2})}\,=\,\sqrt{(2500t^{2})}\,=\,50\left|t\right| \,=\, 50t\,.\]
(time does not assume negative values)
\[v\left(t\right)\,=\, 50\,\mathrm{m \cdot s^{-2}}\cdot t\,.\]
Hint 7 for g): Uniformity of the particle motion
Realize when the movement of the particle is uniform. What holds for its speed?
Solution of Hint 7
The particle performs a uniform motion if its speed is constant.

In our scenario, speed is a linear function of time:
\[v\,=\,50\,\mathrm{m \cdot s^{-2}}\cdot t\,.\]
Therefore movement of the particle is not uniform (it is a case of a uniformly accelerated motion).
Hint 8 for h): Acceleration of the particle
You know the components of the velocity of the particle in directions of the coordinate axes.

How can you use them and the definition of acceleration to obtain the behaviour of the acceleration of the particle?
Solution of Hint 8
From the definition of acceleration we obtain:

\[\vec{a}\left(t\right)\,=\,\frac{\mathrm{d}\vec{v}\left(t\right)}{\mathrm{d}t}\,=\,\frac{\mathrm{d}^{2}\vec{r}\left(t\right)}{\mathrm{d}t^{2}}\,=\,\frac{\mathrm{d}v_\mathrm{x}\left(t\right)}{\mathrm{d}t}\vec{i}+\frac{\mathrm{d}v_\mathrm{y}\left(t\right)}{\mathrm{d}t}\vec{j}+\frac{\mathrm{d}v_\mathrm{z}\left(t\right)}{\mathrm{d}t}\vec{k}\,,\] \[a_\mathrm{x}\,=\,\frac{\mathrm{d}v_\mathrm{x}}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}(30t)\,=\, 30\,,\] \[a_\mathrm{y}\,=\,\frac{\mathrm{d}v_\mathrm{y}}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}\left(-40t\right)\,=\,-40\,, \] \[a_\mathrm{z}\,=\,\frac{\mathrm{d}v_\mathrm{z}}{\mathrm{d}t}\,=\,\frac{\mathrm{d}}{\mathrm{d}t}(0)\,=\,0\,,\] \[\vec{a}\left(t\right)\,=\,30\vec{i}+(-40)\vec{j}+0\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.
Hint 9 for i): Magnitude of acceleration of the particle
You know the behaviour of the acceleration of the particle and its coordinates. How can you determine the behaviour of its magnitude?
Solution of Hint 9
\[a\left(t\right)\,=\,\left|\vec{a}\left(t\right)\right|\,=\,\sqrt{(a_\mathrm{x}^{2}\left(t\right)+a_\mathrm{y}^{2}\left(t\right)+a_\mathrm{z}^{2})}\] \[a\left(t\right)\,=\,\sqrt{\left(900+1600\right)}\,=\,\sqrt{\left(2500\right)}\,=\,50\] \[a\left(t\right)\,=\,50\,\mathrm{m \cdot s^{-2}}\]
Hint 10 for j): Speed and acceleration of the particle when crossing the axes
To calculate the speed of the particle you need to know the relation describing the behaviour of speed and time at which the particle will cross the x axis (or the y axis). You know both from previous Hints (3 and 6).

Proceed similarly to calculate the magnitude of acceleration. Do you need to know the time the particle will cross the axes in this case as well?
Solution of Hint 10
1) The particle crosses the x axis at the moment when:
\[t\,=\,\frac{1}{\sqrt{5}}\,\mathrm{s}\,.\]
From the relation for speed and the magnitude of acceleration we obtain:
\[v\,=\,50t\,,\] \[v\,=\,50(\frac{1}{\sqrt{5}})\,,\] \[v\,=\,10\sqrt{5}\,\mathrm{m \cdot s^{-1}}\,,\] \[a\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]
The magnitude of acceleration is constant, therefore we do not need to know the time when the particle crosses the axis.

2) The particle crosses the y axis at the moment when:
\[t\,=\,0\,\mathrm{s}\,.\]
Similarly to 1):
\[v\,=\,50{\cdot}0\,,\] \[v\,=\,0\,\mathrm{m \cdot s^{-1}}\,,\] \[a\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]
The magnitude of acceleration is constant, therefore we do not need to know the time when the particle crosses the axis.
OVERALL SOLUTION
a)

We can write the position vector as:
\[\vec{r}\left(t\right)\,=\, x\vec{i}+ y\vec{j}+ z\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.

From here:
\[x\left(t\right)\,=\,15t^{2}\,,\] \[y\left(t\right)\,=\,4-20t^{2}\,,\] \[z\left(t\right)\,=\,0\,.\]

b)

The particle crosses the y axis at the moment its x coordinate is equal to 0:
\[x\left(t\right)\,=\,15t^{2}\,,\] \[0\,=\,15t_1^{2}\,,\] \[t_1\,=\,0\,\mathrm{s}\,.\]

c)

The particle crosses the x axis at the moment its y coordinate is equal to 0:
\[y\left(t\right)\,=\,4-20t^{2}\,,\] \[0\,=\,4-20t_2^{2}\,,\] \[t_2^{2}\,=\,\frac{1}{5}\,,\] \[t_2\,=\,\frac{1}{\sqrt{5}}\,\mathrm{s}\,.\]
(The mathematical solution \(t_2\,=\,\frac{-1}{\sqrt{5}}\,\mathrm{s}\) bears no physical significance.)

d)

From the definition of velocity we obtain:

\[\vec{v}(t)\,=\,\frac{\mathrm{d}\vec{r}\left(t\right)}{\mathrm{d}t}\,=\,\frac{\mathrm{d}x\left(t\right)}{\mathrm{d}t}\vec{i}+\frac{\mathrm{d}y\left(t\right)}{\mathrm{d}t}\vec{j}+\frac{\mathrm{d}z\left(t\right)}{\mathrm{d}t}\vec{k}\,,\] \[v_\mathrm{x}\,=\,\frac{\mathrm{d}x}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}\left(15t^{2}\right)\,=\, 30t\,,\] \[v_\mathrm{y}\,=\,\frac{\mathrm{d}y}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}\left(4-20t^{2}\right)\,=\, -40t \,,\] \[v_\mathrm{z}\,=\,\frac{\mathrm{d}z}{\mathrm{d}t}\,=\,\frac{\mathrm{d}}{\mathrm{d}t}\left(0\right)\,=\, 0\,,\] \[\vec{v}\left(t\right)\,=\, 30t\vec{i}+\left(-40t\right)\vec{j}+0\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.

e)

In the direction of x:
\[x\left(t\right)\,=\,15t^{2}\,,\] \[v_x\left(t\right)\,=\, 30t\,.\]
The movement is uniformly accelerated.

In the direction of y:
\[y\left(t\right)\,=\,4-20t^{2}\,,\] \[v_y\left(t\right)\,=\, -40t\,.\]
The movement is uniformly accelerated in the opposite direction.

In the direction of z:
\[z\left(t\right)\,=\, 0\,,\] \[v_z\left(t\right)\,=\, 0\,.\]
The particle is at rest.

f)

From the definition of speed we obtain:

\[v\left(t\right)\,=\,\left|\vec{v}\left(t\right)\right|\,=\,\sqrt{v_\mathrm{x}^{2}\left(t\right)+v_\mathrm{y}^{2}\left(t\right)+v_\mathrm{z}^{2}\left(t\right)}\,,\] \[v\left(t\right)\,=\,\sqrt{(900t^{2}+1600t^{2})}\,=\,\sqrt{(2500t^{2})}\,=\,50\left|t\right| \,=\, 50t\,.\]
(time does not assume negative values)
\[v\left(t\right)\,=\, 50\,\mathrm{m \cdot s^{-2}}\cdot t\,.\]

g)

The particle performs a uniform motion if its speed is constant.

In our scenario, speed is a linear function of time:
\[v\,=\,50\,\mathrm{m \cdot s^{-2}}\cdot t\,.\]
Therefore movement of the particle is not uniform (it is a case of a uniformly accelerated motion).

h)

From the definition of acceleration we obtain:

\[\vec{a}\left(t\right)\,=\,\frac{\mathrm{d}\vec{v}\left(t\right)}{\mathrm{d}t}\,=\,\frac{\mathrm{d}^{2}\vec{r}\left(t\right)}{\mathrm{d}t^{2}}\,=\,\frac{\mathrm{d}v_\mathrm{x}\left(t\right)}{\mathrm{d}t}\vec{i}+\frac{\mathrm{d}v_\mathrm{y}\left(t\right)}{\mathrm{d}t}\vec{j}+\frac{\mathrm{d}v_\mathrm{z}\left(t\right)}{\mathrm{d}t}\vec{k}\,,\] \[a_\mathrm{x}\,=\,\frac{\mathrm{d}v_\mathrm{x}}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}(30t)\,=\, 30\,,\] \[a_\mathrm{y}\,=\,\frac{\mathrm{d}v_\mathrm{y}}{\mathrm{d}t}\,=\, \frac{\mathrm{d}}{\mathrm{d}t}\left(-40t\right)\,=\,-40\,, \] \[a_\mathrm{z}\,=\,\frac{\mathrm{d}v_\mathrm{z}}{\mathrm{d}t}\,=\,\frac{\mathrm{d}}{\mathrm{d}t}(0)\,=\,0\,,\] \[\vec{a}\left(t\right)\,=\,30\vec{i}+(-40)\vec{j}+0\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.

i)
\[a\left(t\right)\,=\,\left|\vec{a}\left(t\right)\right|\,=\,\sqrt{(a_\mathrm{x}^{2}\left(t\right)+a_\mathrm{y}^{2}\left(t\right)+a_\mathrm{z}^{2})}\] \[a\left(t\right)\,=\,\sqrt{\left(900+1600\right)}\,=\,\sqrt{\left(2500\right)}\,=\,50\] \[a\left(t\right)\,=\,50\,\mathrm{m \cdot s^{-2}}\]

j)

1) The particle crosses the x axis at the moment when:
\[t\,=\,\frac{1}{\sqrt{5}}\,\mathrm{s}\,.\]
From the relation for speed and the magnitude of acceleration we obtain:
\[v\,=\,50t\,,\] \[v\,=\,50(\frac{1}{\sqrt{5}})\,,\] \[v\,=\,10\sqrt{5}\,\mathrm{m \cdot s^{-1}}\,,\] \[a\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]
The magnitude of acceleration is constant, therefore we do not need to know the time when the particle crosses the axis.

2) The particle crosses the y axis at the moment when:
\[t\,=\,0\,\mathrm{s}\,.\]
Similarly to 1):
\[v\,=\,50{\cdot}0\,,\] \[v\,=\,0\,\mathrm{m \cdot s^{-1}}\,,\] \[a\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]
The magnitude of acceleration is constant, therefore we do not need to know the time when the particle crosses the axis.
Answer
Note: We do not use units in relations to increase clarity of notation.

a)
\[x\left(t\right)\,=\,15t^{2}\,,\] \[y\left(t\right)\,=\,4-20t^{2}\,,\] \[z\left(t\right)\,=\,0\,.\]

b)

The particle will cross the y axis at time:
\[t_\mathrm{1}\,=\,0\,\mathrm{s}\,.\]

c)

The particle will cross the x axis at time:
\[t_\mathrm{2}\,=\,\frac{1}{\sqrt{5}}\,\mathrm{s}\,\dot{=}\,0{,}45\,\mathrm{s}\,.\]

d)
\[v_\mathrm{x}\left(t\right)\,=\,30t\,,\] \[v_\mathrm{y}\left(t\right)\,=\,-40t\,,\] \[v_\mathrm{z}\left(t\right)\,=\,0\,,\] \[\vec{v}\left(t\right)\,=\,30t\vec{i}+\left(-40t\right)\vec{j}+0\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.

e)

In the direction of x: uniformly accelerated motion

In the direction of y: uniformly accelerated motion in the opposite direction

In the direction of z: the particle is at rest.

f)
\[v\left(t\right)\,=\,50\,\mathrm{m \cdot s^{-2}}t\,.\]

g)

Speed is a linear function of time:
\[v\,=\,50\,\mathrm{m \cdot s^{-2}}\cdot t\,.\]
Therefore the movement of the particle is not uniform (it is a case of uniformly accelerated motion).

h)
\[a_\mathrm{x}\,=\,30\,,\] \[a_\mathrm{y}\,=\,-40\,,\] \[a_\mathrm{z}\,=\,0\,,\] \[\vec{a}\left(t\right)\,=\,30\vec{i}+(-40)\vec{j}+0\vec{k}\,,\]
where \(\vec{i}\), \(\vec{j}\), \(\vec{k}\) are unit vectors in directions of the coordinate axes.

i)
\[a\left(t\right)\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]

j)

1) The particle crosses the x axis at the moment when:

\[t\,=\,\frac{1}{\sqrt{5}}\,\mathrm{s}\,.\]
Tedy:
\[v\,=\,10\sqrt{5}\,\mathrm{m \cdot s^{-1}}\,,\] \[a\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]
2) The particle crosses the y axis at the moment when:
\[t\,=\,0\,\mathrm{s}\,.\]
Therefore:
\[v\,=\,0\,\mathrm{m \cdot s^{-1}}\,,\] \[a\,=\,50\,\mathrm{m \cdot s^{-2}}\,.\]

Difficulty level: Level 4 – Undergraduate level

×Original source: Mandíková, D., Rojko, M.: Soubor úloh z mechaniky pro studium učitelství. I. část. Interní materiál, MFF UK, Praha 1994
Zpracováno v diplomové práci Jany Moltašové (2011).