This page is all about kinematics for vectors. You can describe the position of a moving object using vectors. This is very much like the equation of a straight line except that the parameter is t for time. The equation below is an example: the objects initial position is (2,3) and the velocity is given as the vector \(\left( \begin{matrix} -1 \\ 4 \end{matrix} \right) \).
\(\textbf{r}=\left( \begin{matrix} 2 \\ 3 \end{matrix} \right) +t\left( \begin{matrix} -1 \\ 4 \end{matrix} \right) \)
On this page, you should learn about
- applications of vector lines to kinematics
- speed and velocity
- collisions
The following videos will help you understand all the concepts from this page
This game will help you get a good understand kinematics with vectors. The parameter t is controlled by the slider
- The aim of the game is to fire a torpedo to sink the submarine.
- The submarine is initially positioned at (1,1) and you are at (3, 0)
- Once the submarine is launched, it takes 1 second to calculate and enter the direction of your torpedo shot.
- What direction should you fire your torpedo? Press play to see what happens.
- Change the direction of the torpedo by entering different numbers in red.
- Can you make them collide after 1 second, 2 seconds, 3 seconds?
- How do you make the bomb travel faster?
- You can recreate a different situation if you click ‘change the submarine’.
In the following video we are going to look we will try and gain a conceptual understanding of velocity vectors. One of the key ideas of this topic is to decide if objects collide. It is not enough to consider if their paths cross. We need to think about whether they occupy the same position at the same moment in time.
To get you started, you might like to play this game to give you an idea about what is going on. Try to hit the submarine with the torpedo!
Now let's consider the example below:
A submarine is initially positioned at (0, 5) travels with velocity \(\left( \begin{matrix} 4 \\ -3 \end{matrix} \right) \\ \)ms-1 .
One second later a torpedo is fired from (3, 0) with velocity \(\left( \begin{matrix} 5 \\ 1 \end{matrix} \right) \\ \)ms-1 .
Does the torpedo manage to shoot the submarine?
Notes from the video
Print from here
Here is a quiz that practises the skills from this page
START QUIZ!
The position of an object is given by \(\textbf{r}=\left( \begin{matrix} 2 \\ 3 \end{matrix} \right) +t\left( \begin{matrix} -3 \\ 4 \end{matrix} \right) \)
What is the velocity?
Velocity is a vector
Distances are given in metres and time in seconds
The position of an object is given by \(\textbf{r}=\left( \begin{matrix} -4 \\ 3 \end{matrix} \right) +t\left( \begin{matrix} 5 \\ -12 \end{matrix} \right) \)
What is the speed?
Use Pythagoras' Theorem on velocity vector
speed = \(\sqrt { ({ 5 }^{ 2 }+{ (-12) }^{ 2 } } \)=13ms-1
The object is travelling at 7ms-1 . What is the value of a if a>0
\(\textbf{r}=\left( \begin{matrix} 1\\-2 \\ 1 \end{matrix} \right) +t\left( \begin{matrix} 2 \\ -3\\a \end{matrix} \right) \)
a=
speed = \(\sqrt { { 2 }^{ 2 }+{ (-3) }^{ 2 } +{a}^{2}} \)
\(\sqrt { 13 +{a}^{2}} =7\)
13 + a2 = 49
a2 = 36
a = 6
Which of the following objects can be described with position vector
\(\textbf{r}=\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) +t\left( \begin{matrix} 1 \\ 2 \end{matrix} \right) \)
When t = 0, objects is at (2,1)
Velocity = \(\left( \begin{matrix} 1 \\ 2 \end{matrix} \right) \)
The initial position of the object below is (a,b)
r = 3i + 6j + t(2i - j)
Find a and b
a =
b =
Find a that makes the paths of the two objects parallel
\(\textbf{r}_{A}=\left( \begin{matrix} 0\\-2 \\ 2 \end{matrix} \right) +t\left( \begin{matrix} a \\ 3\\-1 \end{matrix} \right) \) , \(\textbf{r}_{B}=\left( \begin{matrix} 3\\1 \\ 5 \end{matrix} \right) +t\left( \begin{matrix} 6 \\ -9\\3 \end{matrix} \right) \)
a =
For objects must be parallel \(\left( \begin{matrix} a \\ 3\\-1 \end{matrix} \right)=k\left( \begin{matrix} 6 \\ -9\\3 \end{matrix} \right)\\ k=-3\\\)
The two objects are travelling in perpendicular directions
rA = 3i - j +t(ai + 6j)
rB = 2i - 3j +t(3i - 2j)
Find the value of a
a =
Scalar product of the velocities=0
The two objects below pass through the point (2,3). The second one passes the point (2,3) a seconds after the first one.
\(\textbf{r}_{A}=\left( \begin{matrix} 1 \\2 \end{matrix} \right) +t\left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \) , \(\textbf{r}_{B}=\left( \begin{matrix} 12 \\-2 \end{matrix} \right) +t\left( \begin{matrix} -2 \\ 1 \end{matrix} \right) \)
a =
\(\left( \begin{matrix} 2 \\ 3 \end{matrix} \right) =\left( \begin{matrix} 1 \\2 \end{matrix} \right) +1\left( \begin{matrix} 1 \\ 1 \end{matrix} \right) =\left( \begin{matrix} 12 \\-2 \end{matrix} \right) +5\left( \begin{matrix} -2 \\ 1 \end{matrix} \right) \)
A toy airplane sets off from ground level. Its position can be given by
\(\textbf{r}=\left( \begin{matrix} 10\\-8 \\ 0 \end{matrix} \right) +t\left( \begin{matrix} 3 \\ -4\\2 \end{matrix} \right) \)
where distance is measured in metres and time in seconds.
How many seconds before it reaches 15m
The z coordinate gives height
During an air show, two planes, A and B, perform a manoeuvre in which their paths cross in a near miss. The two planes are flying at the same altitude.
\(\textbf{ r }_{ A }=\left( \begin{matrix} 150 \\ 320 \end{matrix} \right) +t\left( \begin{matrix} 200 \\ 300 \end{matrix} \right) \)
\(\textbf{ r }_{ B }=\left( \begin{matrix} 875 \\ 110 \end{matrix} \right) +t\left( \begin{matrix} -100 \\ 400 \end{matrix} \right) \)
t = time in seconds. Distances are given in metres.
a) Show that the two planes cross paths, but the planes do not collide
b) Find the distance between the planes when t = 0.
c) Show that the distance d between A and B at any time t can be given by the expression
d = \(\sqrt { 100000{ t }^{ 2 }-477000t+569725 } \)
d) To the nearest metre, find the closest distance that the two planes get to one another.
Hint
Full Solution
Distances in this question are given in metres
Two brothers, Orville and Wilbur are testing their model airplanes. The position of Orville's airplane t seconds after taking off from ground level is given by
\(\textbf{ r }=\left( \begin{matrix} 12 \\ -19\\0 \end{matrix} \right) +t\left( \begin{matrix} -4 \\ 4\\3 \end{matrix} \right) \)
a) Find the height of the plane after 4 seconds.
b) Wilbur's airplane takes off after Orville's airplane s seconds after taking off is given by
\(\textbf{ r }=\left( \begin{matrix} -26 \\ 25\\0 \end{matrix} \right) +s\left( \begin{matrix} 2 \\ -4\\8 \end{matrix} \right) \)
Find the angle between the two paths.
c) The two airplanes collide at (-20,13,24). How long after Orville’s airplane takes off does Wilbur’s airplane take off?
d) Find the speed of the two airplanes at the moment of the collision.
Hint
Full Solution
All distances in this question are in km.
An interceptor missile, M1 is positioned at the origin. A missile, M2 is launched from (-20,7) with velocity \(\left( \begin{matrix} 3 \\ 1 \end{matrix} \right) \) kms-1 . M1 is capable of twice the speed of M2 . How many seconds later must the interceptor missile, M1 be launched if it is to travel the shortest possible distance?
Hint
Full Solution
How much of Kinematics have you understood?
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