




Conceptual questions
Problems
The acceleration of a particle varies with time according to the equation
$a(t)=p{t}^{2}q{t}^{3}$ . 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?
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Between
t = 0 and
t =
t
_{0} , a rocket moves straight upward with an acceleration given by
$a(t)=AB{t}^{1\phantom{\rule{0.2em}{0ex}}\text{/}2}$ , 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.
$\begin{array}{}\\ \\ v(t)={\displaystyle \int a(t)dt+{C}_{1}}={\displaystyle \int \left(AB{t}^{1\phantom{\rule{0.2em}{0ex}}\text{/}2}\right)dt+{C}_{1}}=At\frac{2}{3}B{t}^{3\phantom{\rule{0.2em}{0ex}}\text{/}2}+{C}_{1}\hfill \\ v(0)=0={C}_{1}\phantom{\rule{0.5em}{0ex}}\text{so}\phantom{\rule{0.5em}{0ex}}v({t}_{0})=A{t}_{0}\frac{2}{3}B{t}_{0}^{\text{3/2}}\hfill \end{array}$ ;
c.
$\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+{C}_{2}={\displaystyle \int \left(At\frac{2}{3}B{t}^{3\phantom{\rule{0.2em}{0ex}}\text{/}2}\right)dt+{C}_{2}}}=\frac{1}{2}A{t}^{2}\frac{4}{15}B{t}^{5\phantom{\rule{0.2em}{0ex}}\text{/}2}+{C}_{2}\hfill \\ x(0)=0={C}_{2}\phantom{\rule{0.5em}{0ex}}\text{so}\phantom{\rule{0.5em}{0ex}}x({t}_{0})=\frac{1}{2}A{t}_{0}^{2}\frac{4}{15}B{t}_{0}^{\text{5/2}}\hfill \end{array}$
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The velocity of a particle moving along the
x axis varies with time according to
$v(t)=A+B{t}^{\mathrm{1}}$ , 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
$x(t=1\phantom{\rule{0.2em}{0ex}}\text{s})=0$ .
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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.
$\begin{array}{}\\ \\ a(t)=3.2{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}t\le 5.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ a(t)=1.5{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}5.0\phantom{\rule{0.2em}{0ex}}\text{s}\le t\le 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ a(t)=0{\text{m/s}}^{2}\phantom{\rule{0.5em}{0ex}}t>11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \end{array}$ ;
b.
$\begin{array}{}\\ \\ x(t)={\displaystyle \int v(t)dt+{C}_{2}={\displaystyle \int 3.2tdt+{C}_{2}=1.6{t}^{2}+{C}_{2}}}\hfill \\ \phantom{\rule{1em}{0ex}}t\le 5.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x(0)=0\Rightarrow {C}_{2}=0\phantom{\rule{0.5em}{0ex}}\text{therefore,}\phantom{\rule{0.2em}{0ex}}x(2.0\phantom{\rule{0.2em}{0ex}}\text{s})=6.4\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ x(t)={\displaystyle \int v(t)dt+{C}_{2}={\displaystyle \int \left[16.01.5\left(t5.0\right)\right]}dt+{C}_{2}}=16t1.5\left(\frac{{t}^{2}}{2}5.0t\right)+{C}_{2}\hfill \\ 5.0\le t\le 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x(5\phantom{\rule{0.2em}{0ex}}\text{s})=1.6{(5.0)}^{2}=40\phantom{\rule{0.2em}{0ex}}\text{m}=16(5.0\phantom{\rule{0.2em}{0ex}}\text{s})1.5\left(\frac{{5}^{2}}{2}5.0\left(5.0\right)\right)+{C}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}40=98.75+{C}_{2}\Rightarrow {C}_{2}=\mathrm{58.75}\hfill \\ x(7.0\phantom{\rule{0.2em}{0ex}}\text{s})=16(7.0)1.5\left(\frac{{7}^{2}}{2}5.0\left(7\right)\right)58.75=69\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1em}{0ex}}x(t)={\displaystyle \int 7.0dt+{C}_{2}=7t+{C}_{2}}\hfill \\ \phantom{\rule{1.5em}{0ex}}t\ge 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \\ x(11.0\phantom{\rule{0.2em}{0ex}}\text{s})=16(11)1.5\left(\frac{{11}^{2}}{2}5.0\left(11\right)\right)58.75=109=7(11.0\phantom{\rule{0.2em}{0ex}}\text{s})+{C}_{2}\Rightarrow {C}_{2}=32\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1em}{0ex}}x(t)=7t+32\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ \phantom{\rule{1.5em}{0ex}}x\ge 11.0\phantom{\rule{0.2em}{0ex}}\text{s}\Rightarrow x(12.0\phantom{\rule{0.2em}{0ex}}\text{s})=7(12)+32=116\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \end{array}$
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Additional problems
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.
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An airplane leaves Chicago and makes the 3000km trip to Los Angeles in 5.0 h. A second plane leaves Chicago onehalf 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:
$\stackrel{\text{\u2013}}{\nu}=600\phantom{\rule{0.2em}{0ex}}\text{km/h}$
2nd plane
$\stackrel{\text{\u2013}}{\nu}=667.0\phantom{\rule{0.2em}{0ex}}\text{km/h}$
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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?
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Questions & Answers
when I click on the links in the topics noting shows. what should I do.
can we regard torque as a force?
Torque is only referred a force to rotate objects.
SHREESH
I need lessons on Simple harmonic motion
Emmanuel
what is the formulae for elastic modulus
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
At time to = 0 the current to the DC motor is reverse, resulting in angular displacement of the motor shafts given by angle = (198rad/s)t  (24rad/s^2)t^2  (2rad/s^3)t^3
At what time is the angular velocity of the motor shaft zero
what is angular velocity
Sadiku
In three experiments, three different horizontal forces are ap
plied to the same block lying on the same countertop. The force
magnitudes are F1 " 12 N, F2 " 8 N, and F3 " 4 N. In each experi
ment, the block remains stationary in spite of the applied force.
Rank the forces according to (a) the
Sadiku
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX
CD=5.83 n direction is NE
Ark
state Hooke's law of elasticity
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded.
F=ke;
Shaibu
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
please help me solve this 👆👆👆
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced.
Weight of water displaced = His weight outside  his weight inside tank
= 720  34.3 = 685.7N
Now, the density of water = 997kg/m³ (this is a known value)
Volume of water displaced = Mass/Density
(next com)
Sharath
density or relative density
Shaibu
Upthrust =72034.3=685.7N
mass of water displayed = 685.7/g
vol of water displayed = 685.7/g/997
hence, density of man = 720/g / (685.7/g/997)
=1046.6 kg/m3
R.d=weight in air/upthrust in water
=720/34.3=20.99
R.d=density of substance/density of water
20.99=x/1
x=20.99g/cm^3
Shaibu
Upthrust = 72034.3=685.7N
vol of water = 685.7/g/density of water = 685.7/g/997
so density of man = 720/g /(685.7/g/997)
=1046.8 kg/m3
is there anyway i can see your calculations
Ian
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
so the density of water is 997
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x>0 but not from the left of x>0
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B)
From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope?
the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
what is the definition of model
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
matthew
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta)
therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆)
on moon, the only difference is the gravity. Gravity on moon = 0.166*g
substituting that value in R, we get the new R
Sharath
Some corrections to my old post.
Time taken to reach ground = 2*v*Sin (∆)/g
R = (2*v²*Sin(∆)*Cos(∆))/g
I put the g in the numerator by mistake in my old post. apologies for that.
R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
a ship move due north at 100kmhr1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result.
V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr.
the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
Kumar
103.077640640442km/h
Peter
state Newton's first law of motion
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia) If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Govindsingh
Source:
OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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