Kinematics: notes and practice questions

If the velocity, $v\ ms^{- 1}$, of a particle, starting from the origin, at time, $t$ seconds, is $v(t) = t^{2}cos(t^{3})$.

$a$ Find the displacement equation.

$b$ Find acceleration after 1 second.

A high-speed drone is being tested, and its acceleration, $a$, is modelled by the function $a(t) = \frac{dv}{dt} = -0.8t + 20$, where $v$ is the speed of the drone in m/s and $t$ is the time in seconds, for $0 \le t \le 30$ seconds.

Determine whether the speed of the drone is increasing or decreasing at $t = 28$ seconds.

It is observed that when $t = 5$ seconds, the speed of the drone is $90$ m/s.

Find an expression for the function $v(t)$.

(a) A drone's vertical velocity, $v$ metres per second, at time $t$ seconds, is given by $v = t \sin(2t^2)$.

Find an expression for the vertical acceleration of the drone.

(b) Hence, or otherwise, find its greatest vertical acceleration for $0 \le t \le 3$ seconds.

(c) The drone starts at ground level (displacement is 0). Find an expression for the vertical displacement of the drone.

(d) Hence show that the drone never descends below ground level.

A drone is launched vertically upwards from a platform. Its vertical velocity, $v \text{ ms}^{-1}$, at time $t$ seconds, is given by the function:

$v = -3t^2 + 18t - 15$, for $t \ge 0$.

Find the times when the drone is momentarily at rest.

Find the magnitude of the drone's vertical acceleration at $t = 5$ seconds.

Find the greatest speed of the drone in the interval $0 \le t \le 5$.

The drone starts from an initial height of $10$ metres above the ground. Find an expression for the height of the drone, $h$ metres, above the ground at time $t$ seconds.

Find the total distance travelled by the drone in the interval $0 \le t \le 4$.

Two autonomous exploration robots, Alpha and Beta, are navigating a straight linear path starting from a central hub, point H. Robot Alpha leaves H at $t = 0$ seconds, and its displacement ($x_A$ metres) from H at time $t$ seconds is given by the equation

$x_A = 6\sqrt{t}$, for $0 \le t \le 9$.

Robot Beta, a newer model, leaves H at $t = 4$ seconds.

After starting its motion, Robot Beta covers any given distance in one-third of the time Robot Alpha would take to cover the same distance.

(a) Write down the equation for the displacement of Beta, $x_B$, in terms of $t$.

(b) Find the value of $t$ at which Robot Beta catches up with Robot Alpha.

(a) The position vector of Drone A at time $t$ seconds is given by $\boldsymbol{r}_A = 4 \cos(3t)\boldsymbol{i} + 5 \sin(3t)\boldsymbol{j}$, where displacement is measured in metres.

(i) Find an expression for the velocity of Drone A at time $t$.

(ii) Hence, find the speed of Drone A when $t = 1.2$ seconds.

(b) (i) Find an expression for the acceleration of Drone A at time $t$.

(ii) Show that the acceleration of Drone A is always directed towards the origin.

(c) The position vector of a second drone, Drone B, is given by $\boldsymbol{r}_B = -5 \sin(4t)\boldsymbol{i} + 4 \cos(4t)\boldsymbol{j}$.

For $0 \le t \le 10$, find the time when the two drones are closest to each other.

(d) At time $k$, where $0 < k < 1.5$, Drone B is moving parallel to Drone A.

(i) Find the value of $k$.

(ii) At time $k$, show that the two drones are moving in the opposite direction.

(a) The profile of a section of a mountain trail can be modelled by the function $y = x^3 - 5x^2 + 2x$, where $y$ is the altitude in hundreds of metres and $x$ is the horizontal distance in hundreds of metres. Find the gradient of the trail at the point where the horizontal distance is $x = 3$ (and the altitude is $y = -12$).

(b) A company's profit, $P$, in thousands of dollars, from selling $x$ units of a new product is modelled by the function $P(x) = 2x^3 - 9x^2 + 12x + 5$. Find the number of units $x$ at which the profit's rate of change is zero. State the corresponding profit for each of these values of $x$.

(c) The concentration of a certain chemical in a solution, $C$, measured in milligrams per litre, is modelled by the function $C(t) = 150 + 20t - 0.5t^2$, where $t$ is the time in minutes after the chemical is added. Find the rate of change of the concentration with respect to time at the instant when $t = 10$ minutes.

(a) A drone launches a rescue package from an initial position of $\begin{pmatrix} 5 \\ 0 \end{pmatrix}$ metres, relative to an origin on the ground. The package is launched with an initial speed of $15\text{ms}^{-1}$ at an angle $\theta$ to the horizontal ground, where $0 < \theta < \frac{\pi}{2}$.

The velocity components of the package, $t$ seconds after it is launched, are given by $v_x(t) = 15\cos\theta$ and $v_y(t) = 15\sin\theta - 9.8t$.

Find an expression for $x$, the horizontal displacement from the origin, in terms of $\theta$ and $t$.

(b) It is given that the vertical displacement of the package from the ground is $y = (15\sin\theta)t - 4.9t^2$. When the package hits the ground, show that $t = \frac{150\sin\theta}{49}$.

(c) Let $x_g$ be the value of $x$ when the package hits the ground.

Find an expression for $x_g$ in terms of $\theta$ only.

(d) Hence, find the value of $\theta$ which maximizes the value of $x_g$.

(e) The model is adapted to account for a horizontal wind with speed $2\text{ms}^{-1}$ acting in the opposite direction to the initial horizontal motion.

(i) In this new model, the horizontal velocity component is $v_x(t) = 15\cos\theta - 2$. The time taken for the package to hit the ground remains $t = \frac{150\sin\theta}{49}$.

Find an expression for $x_g$, the value of $x$ when the package hits the ground, in terms of $\theta$ only.

(ii) Hence, find the value of $\theta$ which maximizes the value of $x_g$.

(a) A drone's displacement, $s$ metres, from a fixed charging station after $t$ minutes (where $t \geq 0$) is given by the equation $s(t) = \frac{t^2}{t+1} + \ln(t+1)$.

Calculate the distance travelled by the drone in the first 3 minutes.

(b) Find an expression for the velocity, $v(t)$, of the drone. Hence, determine if the drone ever becomes stationary for $t \geq 0$.

A robotic system is designed to launch small spherical 'seed pods' into a series of elevated collection bins. The robot is positioned at the origin, O, of a coordinate system on a horizontal platform. In this system, $x$ and $y$ represent the horizontal and vertical displacement from O, measured in metres.

Bin $B_1$ is the closest collection bin to the robot. The coordinates of the centre of bin $B_1$ are $(25, 1.5)$.

Each subsequent bin is $0.8$ m further from O horizontally and $0.4$ m higher than the bin in the row below it. Let bin $B_n$ be the bin in row $n$.

Write down the coordinates of the centre of bin $B_5$.

Find, in terms of $n$, the coordinates for the centre of bin $B_n$.

While in motion, a seed pod can be treated as a projectile.

Let $t$ be the time, in seconds, after a seed pod is launched.

At any time $t > 0$, the acceleration of the seed pod, in m s$^{-2}$, is given by the vector

$\begin{pmatrix} 0 \\ -9.8 \end{pmatrix}$

The initial velocity, in m s$^{-1}$, of the seed pod is given as $\begin{pmatrix} 24.5 \cos \theta \\ 24.5 \sin \theta \end{pmatrix}$, where $\theta$ is the angle to the horizontal at which the seed pod is launched and $0^\circ < \theta \le 90^\circ$.

Find an expression for the velocity, $\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix}$, at time $t$.

Hence show that when the seed pod is launched vertically, the time for it to reach its maximum height is $2.5$ seconds.

The displacement of the seed pod, $t$ seconds after it is launched, is given by the vector equation

$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 24.5(\cos \theta)t \\ 24.5(\sin \theta)t - 4.9t^2 \end{pmatrix}$

Using the given answer to part (b)(ii) or otherwise, find the maximum height reached by a seed pod when it is launched vertically.

If there were no bins to block its path, and the seed pod was launched at an angle $\theta$, show that the value of $x$ when it would hit the ground is given by the expression

$x = \frac{600.25}{9.8} (2 \sin \theta \cos \theta)$.

Hence find the maximum possible value for $x$ if there were no bins to block the path of the seed pod.

In order to calculate which bins can be reached by a seed pod, it is required to find the equation of the curve that forms the boundary of all the points that can be reached. This boundary is represented by a parabolic curve $y = ax^2 + bx + c$, with its vertex $V$ on the $y$-axis.

Using your answers to parts (c) and (d)(ii), or otherwise, find the value of $c$.

Find the value of $b$.

Find the value of $a$.

A technician is considering placing a special sensor in bin $B_{30}$.

Show that it is not possible for a seed pod to ever reach bin $B_{30}$.

The graph below represents the vertical velocity $v$, in metres per second, of a drone over the time interval $t = 0$ to $t = 12$ seconds. Positive velocity indicates upward motion.

At $t = 1\text{s}$, determine whether the drone is moving upwards or downwards.

Find the interval(s) in which the drone is moving i upwards ii downwards.

Explain your answer.

At $t = 6\text{s}$, determine whether the drone's acceleration is positive or negative.

Describe the motion of the drone in each of the intervals $[0,2]$, $[2,5]$, $[5,7]$, $[7,10]$ and $[10,12]$ in terms of its vertical direction, velocity, and acceleration.

Determine the time in the given interval when the drone reaches its maximum height.

A particle is moving through a viscous fluid. Its acceleration, $a\text{ ms}^{-2}$, at time $t\text{ s}$ is given by $a = 6e^{-0.1t}$.

(a) Find an expression for the velocity of the particle, $v\text{ ms}^{-1}$, as a function of time $t\text{ s}$, given that the particle started from rest.

(b) Find an expression for the displacement of the particle, $s\text{ m}$, as a function of time $t\text{ s}$, given that the particle started from the origin.

(c) Write down the limiting velocity of the particle as $t$ gets very large.

(d) Find the particle's displacement 30 seconds after it begins to move.

A drone's acceleration is given by the vector $\mathbf{a}(t) = \begin{pmatrix} 2t \\ 6 \end{pmatrix}$, where $t$ is the time in seconds.

Initially, at $t=0$, the drone is hovering at a position $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ metres and has an initial velocity of $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ m/s.

In terms of $t$, find an expression for the drone's velocity $\mathbf{v}(t)$.

In terms of $t$, find an expression for the drone's displacement $\mathbf{s}(t)$.

When $t = 4$ seconds, find the drone's velocity.

When $t = 4$ seconds, find the drone's displacement.

(a) A weather satellite is in a circular orbit around a planet. Its velocity vector, relative to a fixed ground station, is given by

$\mathbf{v} = \begin{pmatrix} -100\pi \cos(2\pi t) \\ -100\pi \sin(2\pi t) \end{pmatrix}$ m/s

where $t$ is the time in seconds. Distances are measured in metres.

Find its acceleration vector $\mathbf{a} = \begin{pmatrix} \ddot{x}(t) \\ \ddot{y}(t) \end{pmatrix}$.

(b) At time $t=0$ seconds, the satellite is at the position $(0, 150)$ m. Find its displacement vector $\mathbf{s} = \begin{pmatrix} x \\ y \end{pmatrix}$.

(c) When the vertical acceleration of the satellite is zero, find the possible positions for the satellite.

(d) State how many complete orbits the satellite makes in one minute.

A deep-sea exploration probe is launched from a research vessel. Its velocity vector, $\mathbf{v}$, at time $t$ seconds after launch, is given by

$\mathbf{v} = \begin{pmatrix} 2t \\ 3t^2 - 1 \\ 4 \end{pmatrix}$ m/s.

At the moment of launch ($t = 0$), the probe's initial position relative to a fixed origin is $\mathbf{r}(0) = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$ m. Find the position vector, $\mathbf{r}(t)$, of the probe at time $t$.

Find the distance of the probe from the origin when $t = 2$ seconds.

A small drone takes off vertically from a launch pad. Its vertical velocity, $v$, in metres per second, at time $t$ seconds, is given by $v(t) = 4 \cos(t)(1+\sin(t) )$, for $t \ge 0$. The drone next comes to instantaneous rest when $t = a$.

(a) Determine the value of $a$.

(b) Find the maximum vertical velocity of the drone during the interval $0 \le t \le a$.

(c) Calculate the total vertical distance travelled by the drone during the interval $0 \le t \le a$, and hence find its average speed during this interval.

A particle's position, $s$ metres, at time $t$ seconds, is given by the function $s(t)$. For each of the following position functions, determine the velocity function, $v(t)$, and find the instantaneous velocity of the particle at $t = 2$ seconds.

(a) $s(t) = 9$

(b) $s(t) = 3t + 2$

(c) $s(t) = 2t^2 -3t +2$

(d) $s(t) = 2t^3 -2t^2 +2t -5$

(e) $s(t) = \frac{6}{t} +1t -4$

(f) $s(t) = \frac{9}{t} +2t^2 +3$

(a) A drone is ascending vertically. Its velocity, $v$, in m/s, at time $t$ seconds, is given by $v = 0.5t^2 + 2t + 3$ for $t \ge 0$.

Given that the drone starts from the ground (origin), find an expression for its displacement from the ground after $t$ seconds.

(b) Hence, find the distance travelled by the drone during the fifth second of its ascent.