# 3.6 Finding velocity and displacement from acceleration  (Page 3/6)

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## Conceptual questions

When given the acceleration function, what additional information is needed to find the velocity function and position function?

## Problems

The acceleration of a particle varies with time according to the equation $a\left(t\right)=p{t}^{2}-q{t}^{3}$ . Initially, the velocity and position are zero. (a) What is the velocity as a function of time? (b) What is the position as a function of time?

Between t = 0 and t = t 0 , a rocket moves straight upward with an acceleration given by $a\left(t\right)=A-B{t}^{1\phantom{\rule{0.2em}{0ex}}\text{/}2}$ , where A and B are constants. (a) If x is in meters and t is in seconds, what are the units of A and B ? (b) If the rocket starts from rest, how does the velocity vary between t = 0 and t = t 0 ? (c) If its initial position is zero, what is the rocket’s position as a function of time during this same time interval?

a. $A={\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}B={\text{m/s}}^{5\phantom{\rule{0.2em}{0ex}}\text{/}2}$ ;
b. $\begin{array}{}\\ \\ v\left(t\right)=\int a\left(t\right)dt+{C}_{1}=\int \left(A-B{t}^{1\phantom{\rule{0.2em}{0ex}}\text{/}2}\right)dt+{C}_{1}=At-\frac{2}{3}B{t}^{3\phantom{\rule{0.2em}{0ex}}\text{/}2}+{C}_{1}\hfill \\ v\left(0\right)=0={C}_{1}\phantom{\rule{0.5em}{0ex}}\text{so}\phantom{\rule{0.5em}{0ex}}v\left({t}_{0}\right)=A{t}_{0}-\frac{2}{3}B{t}_{0}^{\text{3/2}}\hfill \end{array}$ ;
c. $\begin{array}{}\\ \\ x\left(t\right)=\int v\left(t\right)dt+{C}_{2}=\int \left(At-\frac{2}{3}B{t}^{3\phantom{\rule{0.2em}{0ex}}\text{/}2}\right)dt+{C}_{2}=\frac{1}{2}A{t}^{2}-\frac{4}{15}B{t}^{5\phantom{\rule{0.2em}{0ex}}\text{/}2}+{C}_{2}\hfill \\ x\left(0\right)=0={C}_{2}\phantom{\rule{0.5em}{0ex}}\text{so}\phantom{\rule{0.5em}{0ex}}x\left({t}_{0}\right)=\frac{1}{2}A{t}_{0}^{2}-\frac{4}{15}B{t}_{0}^{\text{5/2}}\hfill \end{array}$

The velocity of a particle moving along the x- axis varies with time according to $v\left(t\right)=A+B{t}^{-1}$ , where A = 2 m/s, B = 0.25 m, and $1.0\phantom{\rule{0.2em}{0ex}}\text{s}\le t\le 8.0\phantom{\rule{0.2em}{0ex}}\text{s}$ . Determine the acceleration and position of the particle at t = 2.0 s and t = 5.0 s. Assume that $x\left(t=1\phantom{\rule{0.2em}{0ex}}\text{s}\right)=0$ .

A particle at rest leaves the origin with its velocity increasing with time according to v ( t ) = 3.2 t m/s. At 5.0 s, the particle’s velocity starts decreasing according to [16.0 – 1.5( t – 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle’s velocity remains constant at 7.0 m/s. (a) What is the acceleration of the particle as a function of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 s, and t = 12.0 s?

a. $\begin{array}{}\\ \\ a\left(t\right)=3.2{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}t\le 5.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ a\left(t\right)=1.5{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}5.0\phantom{\rule{0.2em}{0ex}}\text{s}\le t\le 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ a\left(t\right)=0{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}t>11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \end{array}$ ;
b. $\begin{array}{}\\ \\ x\left(t\right)=\int v\left(t\right)dt+{C}_{2}=\int 3.2tdt+{C}_{2}=1.6{t}^{2}+{C}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}t\le 5.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x\left(0\right)=0⇒{C}_{2}=0\phantom{\rule{0.5em}{0ex}}\text{therefore,}\phantom{\rule{0.2em}{0ex}}x\left(2.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=6.4\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ x\left(t\right)=\int v\left(t\right)dt+{C}_{2}=\int \left[16.0-1.5\left(t-5.0\right)\right]dt+{C}_{2}=16t-1.5\left(\frac{{t}^{2}}{2}-5.0t\right)+{C}_{2}\hfill \\ 5.0\le t\le 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x\left(5\phantom{\rule{0.2em}{0ex}}\text{s}\right)=1.6{\left(5.0\right)}^{2}=40\phantom{\rule{0.2em}{0ex}}\text{m}=16\left(5.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)-1.5\left(\frac{{5}^{2}}{2}-5.0\left(5.0\right)\right)+{C}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}40=98.75+{C}_{2}⇒{C}_{2}=-58.75\hfill \\ x\left(7.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=16\left(7.0\right)-1.5\left(\frac{{7}^{2}}{2}-5.0\left(7\right)\right)-58.75=69\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1em}{0ex}}x\left(t\right)=\int 7.0dt+{C}_{2}=7t+{C}_{2}\hfill \\ \phantom{\rule{1.5em}{0ex}}t\ge 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x\left(11.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=16\left(11\right)-1.5\left(\frac{{11}^{2}}{2}-5.0\left(11\right)\right)-58.75=109=7\left(11.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)+{C}_{2}⇒{C}_{2}=32\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1em}{0ex}}x\left(t\right)=7t+32\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1.5em}{0ex}}x\ge 11.0\phantom{\rule{0.2em}{0ex}}\text{s}⇒x\left(12.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=7\left(12\right)+32=116\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \end{array}$

Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher’s mound? Compare this with the average reaction time of a human to a visual stimulus, which is 0.25 s.

An airplane leaves Chicago and makes the 3000-km trip to Los Angeles in 5.0 h. A second plane leaves Chicago one-half hour later and arrives in Los Angeles at the same time. Compare the average velocities of the two planes. Ignore the curvature of Earth and the difference in altitude between the two cities.

Take west to be the positive direction.
1st plane: $\stackrel{\text{–}}{\nu }=600\phantom{\rule{0.2em}{0ex}}\text{km/h}$
2nd plane $\stackrel{\text{–}}{\nu }=667.0\phantom{\rule{0.2em}{0ex}}\text{km/h}$

Unreasonable Results A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. If his average velocity is 24 km/h, how long did it take him to complete the trip? Is this a reasonable time?

what is electromagnetism
It is the study of the electromagnetic force, one of the four fundamental forces of nature. ... It includes the electric force, which pushes all charged particles, and the magnetic force, which only pushes moving charges.
Energy
what is units?
units as in how
praise
What is th formular for force
F = m x a
Santos
State newton's second law of motion
can u tell me I cant remember
Indigo
force is equal to mass times acceleration
Santos
The acceleration of a system is directly proportional to the and in the same direction as the external force acting on the system and inversely proportional to its mass that is f=ma
David
The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board?
Consider a wave produced on a stretched spring by holding one end and shaking it up and down. Does the wavelength depend on the distance you move your hand up and down?
how can one calculate the value of a given quantity
means?
Manorama
To determine the exact value of a percent of a given quantity we need to express the given percent as fraction and multiply it by the given number.
AMIT
meaning
Winford
briefly discuss rocket in physics
ok let's discuss
Jay
What is physics
physics is the study of natural phenomena with concern with matter and energy and relationships between them
Ibrahim
a potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inductance of the inductor that should be connected to the capacitor for the circuit to oscillate at 1125Hza potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inducta
L= 0.002H
NNAEMEKA
how did you get it?
Favour
is the magnetic field of earth changing
what is thought to be the energy density of multiverse and is the space between universes really space
tibebeab
can you explain it
Guhan
Energy can not either created nor destroyed .therefore who created? and how did it come to existence?
this greatly depend on the kind of energy. for gravitational energy, it is result of the shattering effect violent collision of two black holes on the space-time which caused space time to be disturbed. this is according to recent study on gravitons and gravitational ripple. and many other studies
tibebeab
and not every thing have to pop into existence. and it could have always been there . and some scientists think that energy might have been the only entity in the euclidean(imaginary time T=it) which is time undergone wick rotation.
tibebeab
What is projectile?
An object that is launched from a device
Grant
2 dimensional motion under constant acceleration due to gravity
Awais
Not always 2D Awais
Grant
Awais
why not? a bullet is a projectile, so is a rock I throw
Grant
bullet travel in x and y comment same as rock which is 2 dimensional
Awais
components
Awais
no all pf you are wrong. projectile is any object propelled through space by excretion of a force which cease after launch
tibebeab
for awais, there is no such thing as constant acceleration due to gravity, because gravity change from place to place and from different height
tibebeab
it is the object not the motion or its components
tibebeab
where are body center of mass on present.
on the mid point
Suzana
is the magnetic field of the earth changing?
tibebeab
does shock waves come to effect when in earth's inner atmosphere or can it have an effect on the thermosphere or ionosphere?
tibebeab
and for the question from bal want do you mean human body or just any object in space
tibebeab
A stone is dropped into a well of 19.6m deep and the impact of sound heared after 2.056 second ,find the velocity of sound in air.
9.53 m/s ?
Kyla
In this case, the velocity of sound is 350 m/s.
Zahangir
why?
Kyla
some calculations is need. then you will get exact result.
Zahangir
i mean how? isn't it just a d over t?
Kyla
calculate the time it takes the stone to hit the ground then minus the stone's time to the total time... then divide the total distance by the difference of the time
Snuggly
awit lenard. Hahahah ari ga to!
Kyla