# 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?

how can I convert mile to meter per hour
1 mile * 1609m
Boon
hey can someone show me how to solve the - "Hanging from the ceiling over a baby bed ...." question
i wanted to know the steps
Shrushti
sorry shrushti..
Rashid
which question please write it briefly
Asutosh
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 312 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Note that Mars has an acceleration of gravity of 3.7 m/s2 .
what is summit
Asutosh
highest point on earth
Ngeh
पृथवी को इसके अक्ष पर कितने कोणीय चाल से घूमाऐ कि भूमधय पे आदमी का भार इसके वासतविक भार से 3/5अधिक हो
best
Murari
At a post office, a parcel that is a 20.0-kg box slides down a ramp inclined at 30.0° 30.0° with the horizontal. The coefficient of kinetic friction between the box and plane is 0.0300. (a) Find the acceleration of the box. (b) Find the velocity of the box as it reaches the end of the plane, if the length of the plane is 2 m and the box starts at rest.
As an IT student must I take physics seriously?
yh
Bernice
hii
Raja
IT came from physics and maths so I don't see why you wouldn't
conditions for pure rolling
Md
the time period of jupiter is 11.6 yrs. how far is jupiter from the sun. distance of earth from rhe sun is 1.5*10 to the power 11 meter.
lists 5 drawing instruments and their uses
that is a question you can find on Google, anyway of top of my head, compass, ruler, protractor, try square, triangles.
Rongfang
A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of radius 2R will be
2F
Jacob
it will be doubled
kelvin
double
Devesh
The difference between vector and scaler quantity
vector has both magnitude & direction but scalar has only magnitude
Manash
my marunong ba dto mag prove ng geometry
ron
how do I find resultant of four forces at a point
Inusah
use the socatoa rule
kingsley
draw force diagram, then work out the direction of force.
Rongfang
In a closed system of forces... Summation of forces in any direction or plane is zero... Resolve if there is a need to then add forces in a particular plane or direction.. Say the x direction... Equate it tk zero
Jacob
define moment of inertia
it is the tendency for a body to continue in motion if is or continue to be at rest if it is.
prince
what is Euler s theorem
what is thermocouple?
joining of two wire of different material forming two junctions. If one is hot and another is cold the it will produce emf...
joining of two metal of different materials to form a junction in one is hot & another is cold
Manash
define dimensional analysis
mathematical derivation?
Hira
explain what Newtonian mechanics is.
a system of mechanics based of Newton laws motion this is easy difenation of mean...
Arzoodan
what is the meaning of single term,mechanics?
jyotirmayee
mechanics is the science related to the behavior of physical bodies when some external force is applied to them
Lalita
SO ASK What is Newtonian mechanics in physics? Newtonian physics, also calledNewtonian or classical mechanics, is the description of mechanical events—those that involve forces acting on matter—using the laws of motion and gravitation formulated in the late seventeenth century by English physicist
Suleiman
can any one send me the best reference book for physics?
Prema
concept of physics by HC verma, Fundamentals of Physics, university of physics
tq u.
Prema
these are the best physics books one can fond both theory and applications.
can any one suggest best book for maths with lot of Tricks?
Vivek
what is the water height in barometer?
SUNEELL
13.5*76 cm. because Mercury is 13.5 times dense than Mercury
LOVE
water is 13.5 times dense than the Mercury
LOVE
plz tell me frnds the best reference book for physics along with the names of authors.
Prema
i recomended the reference book for physics from library University of Dublin or library Trinity college
Arzoodan
A little help here... . 1. Newton's laws of Motion, are they applicable to motions of all speeds? 2.state the speeds which are applicable to Newtons laws of Motion
Derek
mechanics which follows Newtons law
Manash
The definition of axial and polar vector .
Arpita
polar vector which have a starting point or pt. of applications is,force,displacement
jyotirmayee
axial vector represent rotational effect and act along the axis of rotation b
jyotirmayee
prove Newton's first law of motion
prince
Hello frnds what is physics in general?
Ngeh
A block of mass m is attached to a spring with spring constant k and free to slide along a horizontal frictionless surface. At t=0, the block spring system is stretched on amount x>0 from the equilibrium position and is released from rest Vx = 0 What is the period of oscillation of the block? What
Ella
What is the velocity of the block when it first comes back to the equilibrium position?
Ella