The problem contains data on the various legs of Jill’s trip, so it would be useful to make a table of the physical quantities. We are given position and time in the wording of the problem so we can calculate the displacements and the elapsed time. We take east to be the positive direction. From this information we can find the total displacement and average velocity. Jill’s home is the starting point
${x}_{0}$ . The following table gives Jill’s time and position in the first two columns, and the displacements are calculated in the third column.
The magnitude of the total displacement is
$\left|\mathrm{-0.75}\right|\phantom{\rule{0.2em}{0ex}}\text{km}=0.75\phantom{\rule{0.2em}{0ex}}\text{km}$ .
The total distance traveled (sum of magnitudes of individual displacements) is
${x}_{\text{Total}}={\displaystyle \sum \left|\text{\Delta}{x}_{\text{i}}\right|=0.5+0.5+1.0+1.75}\phantom{\rule{0.2em}{0ex}}\text{km}=3.75\phantom{\rule{0.2em}{0ex}}\text{km}$ .
We can graph Jill’s position versus time as a useful aid to see the motion; the graph is shown in
[link] .
Significance
Jill’s total displacement is −0.75 km, which means at the end of her trip she ends up
$0.75\phantom{\rule{0.2em}{0ex}}\text{km}$ due west of her home. The average velocity means if someone was to walk due west at
$0.013$ km/min starting at the same time Jill left her home, they both would arrive at the final stopping point at the same time. Note that if Jill were to end her trip at her house, her total displacement would be zero, as well as her average velocity. The total distance traveled during the 58 minutes of elapsed time for her trip is 3.75 km.
Check Your Understanding A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?
(a) The rider’s displacement is
$\text{\Delta}x={x}_{\text{f}}-{x}_{0}=\mathrm{-1}\phantom{\rule{0.2em}{0ex}}\text{km}$ . (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km.
Kinematics is the description of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.
Displacement is the change in position of an object. The SI unit for displacement is the meter. Displacement has direction as well as magnitude.
Distance traveled is the total length of the path traveled between two positions.
Time is measured in terms of change. The time between two position points
${x}_{1}$ and
${x}_{2}$ is
$\text{\Delta}t={t}_{2}-{t}_{1}$ . Elapsed time for an event is
$\text{\Delta}t={t}_{\text{f}}-{t}_{0}$ , where
${t}_{\text{f}}$ is the final time and
${t}_{0}$ is the initial time. The initial time is often taken to be zero.
Average velocity
$\stackrel{\text{\u2013}}{v}$ is defined as displacement divided by elapsed time. If
${x}_{1},{t}_{1}$ and
${x}_{2},{t}_{2}$ are two position time points, the average velocity between these points is
Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Identify each quantity in your example specifically.
You drive your car into town and return to drive past your house to a friend’s house.
Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same?
Bacteria move back and forth using their flagella (structures that look like little tails). Speeds of up to 50 μm/s (50 × 10
^{−6} m/s) have been observed. The total distance traveled by a bacterium is large for its size, whereas its displacement is small. Why is this?
If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.
Consider a coordinate system in which the positive
x axis is directed upward vertically. What are the positions of a particle (a) 5.0 m directly above the origin and (b) 2.0 m below the origin?
A car is 2.0 km west of a traffic light at
t = 0 and 5.0 km east of the light at
t = 6.0 min. Assume the origin of the coordinate system is the light and the positive
x direction is eastward. (a) What are the car’s position vectors at these two times? (b) What is the car’s displacement between 0 min and 6.0 min?
a.
${\overrightarrow{x}}_{1}=(\mathrm{-2.0}\phantom{\rule{0.2em}{0ex}}\text{m})\widehat{i}$ ,
${\overrightarrow{x}}_{2}=(5.0\phantom{\rule{0.2em}{0ex}}\text{m})\widehat{i}$ ; b. 7.0 m east
The Shanghai maglev train connects Longyang Road to Pudong International Airport, a distance of 30 km. The journey takes 8 minutes on average. What is the maglev train’s average velocity?
The position of a particle moving along the
x -axis is given by
$x(t)=4.0-2.0t$ m. (a) At what time does the particle cross the origin? (b) What is the displacement of the particle between
$\text{t}=3.0\phantom{\rule{0.2em}{0ex}}\text{s}$ and
$\text{t}=6.0\phantom{\rule{0.2em}{0ex}}\text{s}?$
a.
$t=2.0$ s; b.
$x(6.0)-x(3.0)=\mathrm{-8.0}-(\mathrm{-2.0})=\mathrm{-6.0}\phantom{\rule{0.2em}{0ex}}\text{m}$
A cyclist rides 8.0 km east for 20 minutes, then he turns and heads west for 8 minutes and 3.2 km. Finally, he rides east for 16 km, which takes 40 minutes. (a) What is the final displacement of the cyclist? (b) What is his average velocity?
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth’s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is 343 m/s at sea level.
a. 150.0 s,
$\stackrel{\text{\u2013}}{v}=156.7\phantom{\rule{0.2em}{0ex}}\text{m/s}$ ; b. 45.7% the speed of sound at sea level
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
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.
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
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
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
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
density
Ian
Upthrust =720-34.3=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
1046.8
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
Kg /cubic meters
how please
Shaibu
Upthrust = 720-34.3=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
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
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
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!
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)
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
what is a vector
smith
a ship move due north at 100kmhr----1 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.
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.