The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north). Velocity is a fundamental concept in kinematics, the branch of classical mechanics that describes the motion of bodies.
Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s or m⋅s−1). For example, "5 metres per second" is a scalar, whereas "5 metres per second east" is a vector. If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration.
This is the problem:
I am a little confused at the solution below. Since ##\vec r_{P/C} = (R/2)*(5 + 4cos(\theta))^{0.5}\ \vec e_r##, I am wondering why there is no vector e_r component for the velocity. I know that d(R)/dt = 0, but you can still get a derivative in the ##\vec e_r## direction...
I have been trying to solve this problem for hours using the mentioned equations but no matter what I do I cannot get the correct answer, that is v = 22.4 m/s. I thought that maybe if I could get an expression where v is a function of time I could solve the problem but I don't know how to do...
To boil down the question, if you have a body at rest and apply a constant force, it will accelerate and the work done on it will be F*s (or the integral version of that statement). However, as the body accelerates due to the force, does that mean, per a given time unit, more and more work will...
Statement of the problem : I copy and paste the statement of the problem to the right as it appeared on the website. Given below is the graph of the ball as its distance from a fixed point with time.
Attempt : Where does this fixed point, say ##\text{P}## lie?
Imagine the fixed point lied...
The correct answer is u=vcos\theta. I have understood so far to be able to conclude that \text{displacement of string} = PA - PC \approx AB
Also, \overline{AB}=\overline{AC}cos\theta
or, more generally, \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta
Now, I had hoped...
position and velocity expressions for the initial conditions
1) x(0)=Acosφ = xi
2) v(0)=-ωAsinφ= vi
Dividing 2nd equation by equation results in:
-ωAsinφ/Acosφ = xi/vi
tanφ = xi/(-ωvi)
tanφ = 0.4
And here they (the textbook) got φ = 0.127π, This is where I...
Here is my attempt. At ymax the velocity turn to zero so we get time t*=v0/g and ymax=1/2 (v0^2/g). At the height y max, since the velocity at this point is 0, i get another equation y= 1/2(v0^2/g)-(g/2)t^2, this equation could be considered as continuation of first equation. Set ymax/2=1/4...
Since space is curved within the Earth's gravitational field, every body that moves there will follow the curvature of space no matter what speed it has, so what will its trajectory be, how will it be straight, only if the launch is made absolutely vertically towards sea level? Only then can it...
Here is the exercise:
Which one seems logical and correct ?
this one:
[ Normally when we increase distances, the velocity of sound decreases?]
Or this one?
You might wonder why. Well, my friend in class told me that the second table could be correct because the experiment was conducted at...
the answer to this question uses the above formula with the tangent function and solves for the initial velocity,
i used the equation (v.sinθ^2) = (v.sinθ)^2 - 2gΔy, setting final velocity equal to zero and solving for initial velocity. this kinematic equation gives a different answer. can...
Hello everyone,
I'm reading Morins book which I like, and I feel I kind of understand the part on time dilation, however I'm a little confused by the geometry of the Pythagorian theorem when applied to velocities.
On the moving clock he shows the velocity of light on the diagonal it traces...
I understand that through process of elimination the only plausible solution is (E), but a question that rises up:
When the ball bounces, does the velocity change from negative to positive instantly (as shown by the dotted lines) or gradually (a very small time period, but still solid line)?
Problem statement : I copy and paste the problem as it appears in the text down below. I have only changed the symbol of the given acceleration from ##a\rightarrow a_0##, owing to its constancy.
Attempt : I must admit that I could proceed very little.
Given...
1. The first equation between velocity ##v## and time ##t## can be derived using the graph I have drawn for the purpose as shown on the right. Since acceleration ##a_0## is a constant, the graph of ##v-t## is a straight line. The slope of the line is ##\dfrac{v-v_0}{t} = a_0\Rightarrow \boxed{v...
Hi,
I am having problems with task b
I then defined the velocity vector and the acceleration vector as follows
##dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##
and
##ddot{\textbf{r}}'(t) =...
My solution was that the final velocity of the fly is equal to the mass of the elephant divided by the mass of the fly, and then multiplies by the delta in the elephant's velocity. My teacher said it was the wrong answer and that the calculations are presumably pretty long
so then when we model the velocity of the motion of a rocket (e.g., in a mathematics report), is it reasonable to assume that the initial velocity equals zero? I also apologise for my lack of information if I made a huge mistake in my question 🙏🏻
Here is only my solution:
##A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}##,
so by integrating we get
##h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.##
Setting ##h(T)=0## we get
##T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.##
By doing the first time derivative of ##h## we...
First, I have always consider that the angular momentum equals to inertia times angular velocity, but that’s not the case from the options perpective, is my memory wrong, or is there something wrong with the options?
Another, I think I need to figure out the angle it went through, I think it has...
Given that the ions are initially at rest my initial velocity is 0. Therefore my Vavg is equal to vf/2
Using the formula Vavg = Change in positon/time, I can solve vf to be equal to 2r/t.
Using the momentum principle, I get an equation of 2r/t = FnetT/12m -> Given that the mass of the ion is...
I think it's not possible.
In 3D for constant acceleration we have : ##\vec v = \vec v_0 + \vec a t##
It's a line in 3 dimension so velocity's magnitude(speed) is changing with time.
I appreciate any better idea.
TL;DR Summary: Almost as difficult as the Physics Olympics semi-finals or finals
The coefficient of friction of the two arcs is the same.The radius of the ball is not taken into account.The roll of the ball is ignored.Can it be calculated or estimated quantitatively?The difficulty is estimated...
EDIT: I've finally found the solution, so here's what I did.
First, calculate Work using the equation: (F-mgsin(theta))*displacement, where F=the force being applied by the push, theta is the angle of the ramp, and displacement is the length of the ramp.
Now that you have the value for work...
800 - (32 x 9.8) = 32v/0.18 where v = velocity
this gives me v = 2.736 m/s
The answer given, however, is 800 = 32v/0.18, i.e. v = 4.5 m/s
The difference, of course, is the weight of the child. I don't understand why this is not allowed for in the net force acting on the child. Can someone put me...
initial velocity y component is (cos30) * 20.
t = 25m / ((cos30) * 20)m/s = 1.45 seconds
d = vt + .5at^2
v= 20sin30
v= 10 , d= 10(1.45s) + .5(-9.8m/s^s)(1.45s)^2
d=4.2m
4.2-2.5 = +1.7m, so the ball will not hit the fence
I need confirmation please
is it possible?
I've given it quite some thought, and my conclusion was that it would be possible, cuz right after I jump, I'll still have the same velocity in the same direction as the running train?
I am going through https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf, here I don't understand how we wrote eq. 1.5.3 with the help of three-vectors to get eq. 1.5.4.
From the bus driver's point of view, who is at rest, the ball's initial velocity is ##u+v##. After the collision, its velocity has to have the same value, but an opposite direction, so ##-(u+v)##. So that means that relative to me standing on the ground at rest, the ball's new velocity is...
1) Using "The person catches the ball at exactly the same height it was thrown from.", we can isolate t by solving yb(t) = v0*sin(theta)*t - (1/2)*g*(t^2) = 0:
yb(t) = v0*sin(theta)*t - (1/2)*g*(t^2) = 0
v0*sin(theta)*t = (1/2)*g*(t^2)
2*v0*sin(theta) = g*t
t = 2*v0*sin(theta) / g
2) At the...
Is it possible to trace the trajectory of an object using only its velocity and position, both of which are given as components. My method of doing so involves using the time until max height is reached, and using that time value to calculate the max height itself (h,k), then plugging in the...
u is the speed of the message relative to the station
v is the speed of the spaceship relative to the station
u’ is the speed of the message relative to the spaceship
u=(-0.5c+0.7c)/(1+((-0.5c)(0.7c))/c^2 )
=0.2c/0.65=0.308c
This just seems way too high, and I'm not sure if I'm doing it...
This is a famous book in India. I was wondering if one could say if the answer should include velocity or speed. I mean, I don't think there are any details which hint at velocity. We are gives speed in the question and we are asked to find out the distance traveled, this hints we are asked to...
I'm not sure.
A source said New Velocity = Current Velocity - TerminalVelocity every second until New Velocity = TerminalVelocity
But it doesn't seem right
edit: I don't know why my latex isn't rendering, any help would be appreciated.
Edit 2: The question was due to a misunderstanding I had, I thought integrating instantaneous velocity would give me average velocity.
I have attached what I have tried so far. I had a doubt. Can you calculate the...
TL;DR Summary: Velocity and the human experience
As I was walking into work this morning I looked up and noticed an airplane flying overhead. Airplanes travel at speeds of up to 550 miles per hour while in flight. This question is really less about physics and more about our everyday...
On many websites etc the velocity of a freefalling object in a vacuum is shown as follows.
After 1,2,3 and 4 seconds respectively;
9.8m/s; 19.6m/s; 29.4m/s; 39.2m/s
I worked out the distance travelled by a freefalling object in a vacuum using d = at^2/2, or, d = 0.5gt^2 and got, for after...
For this,
Does someone please know whether they assume for the equation highlighted that ##\frac{v}{f} ≥ \frac{v_S}{f}## since otherwise the wavelength would be negative (which I assume is impossible)?
Many thanks!
Hello Physicsforum!
This is my attempt:
First I realised:
##a_s=a_n##
Secondly I used since previus known formulas:
##a_n=\frac {v^2} {R}##
##v=v_0+a_s*t##
Although now I do not know how to continue, any suggestions would be appriciated!
Thanks for your help on beforehand :smile:
I did try to solve the problem by forming the derivative and my result was: v=(rcos(phi), -rsin(phi),0). My solution is wrong, the tutor corrected the task but he didn’t give us the results. My question is what the solution is. Thanks in advance.
The question was this:
My calculations show that the answer should be equal to work done on crate to make it reach the same velocity which is equal to 216 J but the answer given is 432 J
It is believed that extra energy is needed to overcome friction but friction is an internal force and...