# 4.2 Acceleration vector  (Page 3/4)

 Page 3 / 4

## Solution

(a) The origin of the coordinate system is at the top of the hill with y- axis vertically upward and the x- axis horizontal. By looking at the trajectory of the skier, the x- component of the acceleration is positive and the y- component is negative. Since the angle is $15\text{°}$ down the slope, we find

${a}_{x}=\left(2.1\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\phantom{\rule{0.2em}{0ex}}\text{cos}\left(15\text{°}\right)=2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}$
${a}_{y}=\left(-2.1\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}15\text{°}=-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}.$

Inserting the initial position and velocity into [link] and [link] for x , we have

$x\left(t\right)=75.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(4.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)t+\frac{1}{2}\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){t}^{2}$
${v}_{x}\left(t\right)=4.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)t.$

For y , we have

$y\left(t\right)=-50.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)t+\frac{1}{2}\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){t}^{2}$
${v}_{y}\left(t\right)=-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)t.$

(b) Now that we have the equations of motion for x and y as functions of time, we can evaluate them at t = 10.0 s:

$x\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=75.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(4.1\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)+\frac{1}{2}\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}^{2}=216.0\phantom{\rule{0.2em}{0ex}}\text{m}$
${v}_{x}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=4.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=24.1\text{m}\text{/s}$
$y\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-50.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)+\frac{1}{2}\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}^{2}=-88.0\phantom{\rule{0.2em}{0ex}}\text{m}$
${v}_{y}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-6.5\phantom{\rule{0.2em}{0ex}}\text{m/s}.$

The position and velocity at t = 10.0 s are, finally,

$\stackrel{\to }{r}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left(216.0\stackrel{^}{i}-88.0\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}\text{m}$
$\stackrel{\to }{v}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left(24.1\stackrel{^}{i}-6.5\stackrel{^}{j}\right)\text{m/s}.$

The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h.

## Significance

It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we can find the position, velocity, and acceleration at any later time.

With [link] through [link] we have completed the set of expressions for the position, velocity, and acceleration of an object moving in two or three dimensions. If the trajectories of the objects look something like the “Red Arrows” in the opening picture for the chapter, then the expressions for the position, velocity, and acceleration can be quite complicated. In the sections to follow we examine two special cases of motion in two and three dimensions by looking at projectile motion and circular motion.

At this University of Colorado Boulder website , you can explore the position velocity and acceleration of a ladybug with an interactive simulation that allows you to change these parameters.

## Summary

• In two and three dimensions, the acceleration vector can have an arbitrary direction and does not necessarily point along a given component of the velocity.
• The instantaneous acceleration is produced by a change in velocity taken over a very short (infinitesimal) time period. Instantaneous acceleration is a vector in two or three dimensions. It is found by taking the derivative of the velocity function with respect to time.
• In three dimensions, acceleration $\stackrel{\to }{a}\left(t\right)$ can be written as a vector sum of the one-dimensional accelerations ${a}_{x}\left(t\right),{a}_{y}\left(t\right),\text{and}\phantom{\rule{0.2em}{0ex}}{a}_{z}\left(t\right)$ along the x- , y -, and z- axes.
• The kinematic equations for constant acceleration can be written as the vector sum of the constant acceleration equations in the x , y , and z directions.

## Conceptual questions

If the position function of a particle is a linear function of time, what can be said about its acceleration?

If an object has a constant x -component of the velocity and suddenly experiences an acceleration in the y direction, does the x- component of its velocity change?

No, motions in perpendicular directions are independent.

If an object has a constant x- component of velocity and suddenly experiences an acceleration at an angle of $70\text{°}$ in the x direction, does the x- component of velocity change?

## Problems

The position of a particle is $\stackrel{\to }{r}\left(t\right)=\left(3.0{t}^{2}\stackrel{^}{i}+5.0\stackrel{^}{j}-6.0t\stackrel{^}{k}\right)\phantom{\rule{0.2em}{0ex}}\text{m}.$ (a) Determine its velocity and acceleration as functions of time. (b) What are its velocity and acceleration at time t = 0?

A particle’s acceleration is $\left(4.0\stackrel{^}{i}+3.0\stackrel{^}{j}\right)\text{m/}{\text{s}}^{2}.$ At t = 0, its position and velocity are zero. (a) What are the particle’s position and velocity as functions of time? (b) Find the equation of the path of the particle. Draw the x- and y- axes and sketch the trajectory of the particle.

a. $\stackrel{\to }{v}\left(t\right)=\left(4.0t\stackrel{^}{i}+3.0t\stackrel{^}{j}\right)\text{m/s},$ $\stackrel{\to }{r}\left(t\right)=\left(2.0{t}^{2}\stackrel{^}{i}+\frac{3}{2}{t}^{2}\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}\text{m}$ ,
b. $x\left(t\right)=2.0{t}^{2}\text{m,}\phantom{\rule{0.2em}{0ex}}y\left(t\right)=\frac{3}{2}{t}^{2}\text{m,}\phantom{\rule{0.2em}{0ex}}{t}^{2}=\frac{x}{2}⇒y=\frac{3}{4}x$

A boat leaves the dock at t = 0 and heads out into a lake with an acceleration of $2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\stackrel{^}{i}.$ A strong wind is pushing the boat, giving it an additional velocity of $2.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\stackrel{^}{i}+1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\stackrel{^}{j}.$ (a) What is the velocity of the boat at t = 10 s? (b) What is the position of the boat at t = 10s? Draw a sketch of the boat’s trajectory and position at t = 10 s, showing the x- and y -axes.

The position of a particle for t >0 is given by $\stackrel{\to }{r}\left(t\right)=\left(3.0{t}^{2}\stackrel{^}{i}-7.0{t}^{3}\stackrel{^}{j}-5.0{t}^{-2}\stackrel{^}{k}\right)\phantom{\rule{0.2em}{0ex}}\text{m}.$ (a) What is the velocity as a function of time? (b) What is the acceleration as a function of time? (c) What is the particle’s velocity at t = 2.0 s? (d) What is its speed at t = 1.0 s and t = 3.0 s? (e) What is the average velocity between t = 1.0 s and t = 2.0 s?

a. $\stackrel{\to }{v}\left(t\right)=\left(6.0t\stackrel{^}{i}-21.0{t}^{2}\stackrel{^}{j}+10.0{t}^{-3}\stackrel{^}{k}\right)\text{m/s}$ ,
b. $\stackrel{\to }{a}\left(t\right)=\left(6.0\stackrel{^}{i}-42.0t\stackrel{^}{j}-30{t}^{-4}\stackrel{^}{k}\right)\text{m/}{\text{s}}^{2}$ ,
c. $\stackrel{\to }{v}\left(2.0s\right)=\left(12.0\stackrel{^}{i}-84.0\stackrel{^}{j}+1.25\stackrel{^}{k}\right)\text{m/s}$ ,
d. $\stackrel{\to }{v}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=6.0\stackrel{^}{i}-21.0\stackrel{^}{j}+10.0\stackrel{^}{k}\text{m/s},\phantom{\rule{0.2em}{0ex}}|\stackrel{\to }{v}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=24.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
$\stackrel{\to }{v}\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=18.0\stackrel{^}{i}-189.0\stackrel{^}{j}+0.37\stackrel{^}{k}\text{m/s},$ $|\stackrel{\to }{v}\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=199.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ ,
e. $\stackrel{\to }{r}\left(t\right)=\left(3.0{t}^{2}\stackrel{^}{i}-7.0{t}^{3}\stackrel{^}{j}-5.0{t}^{-2}\stackrel{^}{k}\right)\text{cm}$
$\begin{array}{}\\ \\ \hfill {\stackrel{\to }{v}}_{\text{avg}}& =9.0\stackrel{^}{i}-49.0\stackrel{^}{j}-6.3\stackrel{^}{k}\text{m/s}\hfill \end{array}$

The acceleration of a particle is a constant. At t = 0 the velocity of the particle is $\left(10\stackrel{^}{i}+20\stackrel{^}{j}\right)\text{m/s}.$ At t = 4 s the velocity is $10\stackrel{^}{j}\text{m/s}.$ (a) What is the particle’s acceleration? (b) How do the position and velocity vary with time? Assume the particle is initially at the origin.

A particle has a position function $\stackrel{\to }{r}\left(t\right)=\text{cos}\left(1.0t\right)\stackrel{^}{i}+\text{sin}\left(1.0t\right)\stackrel{^}{j}+t\stackrel{^}{k},$ where the arguments of the cosine and sine functions are in radians. (a) What is the velocity vector? (b) What is the acceleration vector?

a. $\stackrel{\to }{v}\left(t\right)=\text{−sin}\left(1.0t\right)\stackrel{^}{i}+\text{cos}\left(1.0t\right)\stackrel{^}{j}+\stackrel{^}{k}$ , b. $\stackrel{\to }{a}\left(t\right)=\text{−cos}\left(1.0t\right)\stackrel{^}{i}-\text{sin}\left(1.0t\right)\stackrel{^}{j}$

A Lockheed Martin F-35 II Lighting jet takes off from an aircraft carrier with a runway length of 90 m and a takeoff speed 70 m/s at the end of the runway. Jets are catapulted into airspace from the deck of an aircraft carrier with two sources of propulsion: the jet propulsion and the catapult. At the point of leaving the deck of the aircraft carrier, the F-35’s acceleration decreases to a constant acceleration of $5.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}$ at $30\text{°}$ with respect to the horizontal. (a) What is the initial acceleration of the F-35 on the deck of the aircraft carrier to make it airborne? (b) Write the position and velocity of the F-35 in unit vector notation from the point it leaves the deck of the aircraft carrier. (c) At what altitude is the fighter 5.0 s after it leaves the deck of the aircraft carrier? (d) What is its velocity and speed at this time? (e) How far has it traveled horizontally?

what is physics
Physics is the tool humans use to understand the properties characteristics and interactions of where they live - the universe. Thus making laws and theories about the universe in a mathematical way derived from empirical results yielded in tons of experiments.
Jomari
This tool, the physics, also enhances their way of thinking. Evolving integrating and enhancing their critical logical rational and philosophical thinking since the greeks fired the first neurons of physics.
Jomari
nice
Satyabrata
Physics is also under the category of Physical Science which deals with the behavior and properties of physical quantities around us.
Angelo
Physical Science is under the category of Physics*... I prefer the most is Theoretical Physics where it deals with the philosophical view of our world.
Jomari
what is unit
Metric unit
Arzoodan
A unit is what comes after a number that gives a precise detail on what the number means. For example, 10 kilograms, 10 is the number while "kilogram" is the unit.
Angelo
there are also different types of units, but metric is the most widely used. It is called the SI system. Please research this on google.
Angelo
Unit? Bahay yon
Jomari
How did you get the value as Dcd=0.2Dab
Why as Dcd=0.2Dab? where are you got this formula?...
Arzoodan
since the distance Dcd=1.2 and the distance Dab=6.0 the ratio 1.2/6.0 gives the equation Dcd=0.2Dab
sunday
Well done.
Arzoodan
how do we add or deduct zero errors from result gotten using vernier calliper?
how can i understand if the function are odd or even or neither odd or even
hamzaani
I don't get... do you mean positive or negative@hamzaani
Aina
Verner calliper is an old calculator
Antonio
Function is even if f(-x) =f(x)
Antonio
Function is odd if f(-x) = - f(x)
Antonio
what physical phenomena is resonance?
is there any resonance in weight?
amrit
Resonance is due to vibrations and waves
Antonio
wait there is a chat here
dare
what is the difference between average velocity and magnitude of displacement
ibrahim
how velocity change with time
ibrahim
average velocity can be zero positive negative but magnitude of displacement is positive
amrit
if there is different displacement in same interval of time
amrit
Displacement can be zero, if you came back
Antonio
Displacement its a [L]
Antonio
Velocity its a vector
Antonio
Speed its the magnitude of velocity
Antonio
[Vt2-Vt1]/[t2-t1] = average velocity,another vector
Antonio
Distance, that and only that can't be negative, and is not a vector
Antonio
Distance its a metrical characteristic of the euclidean space
Antonio
Velocity change in time due a force acting (an acceleration)
Antonio
the change in velocity can be found using conservation of energy if the displacement is known
Jose
BEFORE = AFTER
Jose
kinetic energy + potential energy is equal to the kinetic energy after
Jose
the potential energy can be described as made times displacement times acceleration. I.e the work done on the object
Jose
mass*
Jose
from there make the final velocity the subject and solve
Jose
If its a conservative field
Antonio
So, no frictions in this case
Antonio
right
Jose
and if still conservative but force is in play then simply include work done by friction
Jose
Is not simple, is a very unknown force
Antonio
the vibration of a particle due to vibration of a similar particle close to it.
Aina
No, not so simple
Antonio
Frequency is involved
Antonio
mechanical wave?
Aina
All kind of waves, even in the sea
Antonio
will the LCR circut pure inductive if applied frequency becomes more than the natural frequency of AC circut? if yes , why?
LCR pure inductive? Is an nonsense
Antonio
what is photon
Photon is the effect of the Maxwell equations, it's the graviton of the electromagnetic field
Antonio
a particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.
Areej
Quantum it's not exact, its the elementary particle of electromagnetic field. Its not well clear if quantum theory its so, or if it's classical mechanics improved
Antonio
A photon is first and foremost a particle. And hence obeys Newtonian Mechanics. It is what visible light and other electromagnetic waves is made up of.
eli
No a photon has speed of light, and no mass, so is not Newtonian Mechanics
Antonio
photon is both a particle and a wave (It is the property called particle-wave duality). It is nearly massless, and travels at speed c. It interacts with and carries electromagnetic force.
Angelo
what are free vectors
a vector hows point of action doesn't static . then vector can move bodily from one point to another point located on its original tragectory.
Anuj
A free vector its an element of an Affine Space
Antonio
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at 45° with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at 90° with respect to the 50-yard line for 12 m, with an elapsed time of 1.2 s. (a) What is Matthews’ final displacement from the start of the play? (b) What is his average velocity?
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at 45Â° with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at 90Â° with respect to the 50-yard line for 12 m, with an elap
ibrahim
Very easy man
Antonio
how to find time moved by a mass on a spring
Maybe you mean frequency
Antonio
why hot soup is more tastier than cold soup?
energy is involved
michael
hot soup is more energetic and thus enhances the flavor than a cold one.
Angelo
Its not Physics... Firstly, It falls under Anatomy. Your taste buds are the one to be blame not its coldness or hotness. Secondly, it depends on how the soup is done. Different soups possess different flavors and savors. If its on Physics, coldness of the soup will just bore you and if its hot...
Jomari
what is the importance of banking road in the circular path
the coefficient of static friction of the tires and the pavement becomes less important because the angle of the banked curve helps friction to prevent slipping
Jose
an insect is at the end of the ring and the ring is rotating at an angular speed 'w' and it reaches to centre find its angular speed.
Angular speed is the rate at which an object changes its angle (measured) in radians, in a given time period. Angular speed has a magnitude (a value) only.  v represents the linear speed of a rotating object, r its radius, and ω its angular velocity in units of radians per unit of time, then v = rω
Angular speed = (final angle) - (initial angle) / time = change in position/time. ω = θ /t. ω = angular speed in radians/sec.
a boy through a ball with minimum velocity of 60 m/s and the ball reach ground 300 metre from him calculate angle of inclination
what is the fomula for work done
work= force x distance
Guest
force × distance
Akash
Foece and displacement along the same direction as that of the force
nalin
force×displacement×cos∆ where ∆ is the angle between displacement and force.....i.e dot product of force and displacement
Is the angle between direcrion and force...
Arzoodan
Work is F x d = [F] •[d] • cos(a°)
Antonio
Force × distance along the same plane....
Aina