In mechanics, acceleration is the rate of change of the velocity of an object with respect to time.
Accelerations are vector quantities (in that they have magnitude and direction). The orientation of an object's acceleration is given by the orientation of the net force acting on that object. The magnitude of an object's acceleration, as described by Newton's Second Law, is the combined effect of two causes:
the net balance of all external forces acting onto that object — magnitude is directly proportional to this net resulting force;
that object's mass, depending on the materials out of which it is made — magnitude is inversely proportional to the object's mass.The SI unit for acceleration is metre per second squared (m⋅s−2,
m
s
2
{\displaystyle {\tfrac {\operatorname {m} }{\operatorname {s} ^{2}}}}
).
For example, when a vehicle starts from a standstill (zero velocity, in an inertial frame of reference) and travels in a straight line at increasing speeds, it is accelerating in the direction of travel. If the vehicle turns, an acceleration occurs toward the new direction and changes its motion vector. The acceleration of the vehicle in its current direction of motion is called a linear (or tangential during circular motions) acceleration, the reaction to which the passengers on board experience as a force pushing them back into their seats. When changing direction, the effecting acceleration is called radial (or orthogonal during circular motions) acceleration, the reaction to which the passengers experience as a centrifugal force. If the speed of the vehicle decreases, this is an acceleration in the opposite direction and mathematically a negative, sometimes called deceleration, and passengers experience the reaction to deceleration as an inertial force pushing them forward. Such negative accelerations are often achieved by retrorocket burning in spacecraft. Both acceleration and deceleration are treated the same, they are both changes in velocity. Each of these accelerations (tangential, radial, deceleration) is felt by passengers until their relative (differential) velocity are neutralized in reference to the vehicle.
Just started learning about uniform circular motion. I really don't understand how we get aΔt2/2 on the side. I also searched on the internet for a similar derivation, but there are none so simple.
Thanks for your help!
P.S There is a mistake in calculation in second line (textbook error).
I'm working on a project where we have a mass (50 kg) sitting on a spring (350 N/mm) and are subjecting it to a sudden impulse (20g) along the spring axis to simulate a shock. We have the profile of the acceleration defined as:
##a(t) = x''(t) = P\cdot \sin^2 (\pi \cdot t / T)##
Where P (peak...
In the case where ground clearence isn't the limiting factor;
Since motorycles are steered using countersteering, a motorcycle cornering at constant max g can achieve any decrease of lean angle (bringing the bike up) by turning harder towards the inside of the turn momentarily, which is not...
Therefore, if someone were to ask what the magnitude of centripetal acceleration is at the top of the wheel at a given instant (relative to the ground):
##v_{cm} = v_{translational, center-of-mass/wheel}##
##ω = ω_{point-of-contact}##
##v_{top} = 2(v_{cm}) = 2(rω)##
##a_{c(top)} =...
Since the acceleration ##\vec a## is given by ##\vec a = \frac{d^2 \vec x}{dt^2}##, it is a function of ##t## only. Of course, the derivative implies that ##t = t(\vec x)## so we can also in principle express ##\vec a## in terms of ##\vec x##. But how can we express an acceleration dependent on...
Because the friction is the same in both parts, the calculated acceleration from (b) should be the same for (c)
I knew I could find Vf, and thought I could do it with an energy equation
Ei=Ef
mgh=1/2mv^2
gh=1/2v^2
(2)9.81(1.5)=1/2v^2(2)
(square root)29.43=(square root)v^2
v= 5.424
Then...
We are given that ##v' = \frac{1}{10}v^2 - g##.
I tried using implicit differentiation so that ##v'' = \frac{1}{5}vv' = \frac{1}{5}v(\frac{1}{10}v^2-g)## and set this equal to 0. Hence we have 3 critical points, at ##v= 0##, and ##v = \pm \sqrt{10g}##.
Calculating ##v''(0)=-120##, we know the...
I was just reading a set of thermodynamics lecture notes and came across the following
In most thermodynamics problems I have done, it is indeed assumed that the piston does not accelerate so we can simply equate forces on the piston. However, I don't fully understand the line of reasoning...
Initially we are given the statement Vav = (x-x0)/t, so far so good. But, we encounter the following paragraph...
"We can also get a second expression for Vav that is valid only when the acceleration is constant, so that the v-t graph is a straight line (as in Fig 2-14 - [I've omitted the graph...
Hello, I hope you are all having a great day !
I've got a physics test in a couple of days and I have some questions:1.
In a calculation, if the acceleration is in m/s², I presume the speed also has to be in m/s and not in km/h ?
2.
So with this graph (v with t), I have to find the total...
Well, I just had this thought earlier, and I want to share it. Here it is.
So, we all know about inertia, right? The resistance to acceleration, or change in motion. Well, there is also a concept about derivatives of acceleration, mainly jerk and yank. If you don't know, jerk is said to be the...
So I figured out the equation, but it is probably wrong because the answer doesn't tally.
Since the string is inextensible, I can assume that tension is the same for both sides, and acceleration for both masses is the same too So:
I can say that the acceleration of 2kg block =acceleration of 7kg...
a)
Eg = Gme/r^2
r = √Gme/Eg
r = √[(6.67x10^-11 N*m^2*kg^2)(5.98x10^24 kg)]/(4.5 N/kg)
r = 9.41x10^6 m
h = r2 - r1
h = 9.41x10^6 m - 6.38x10^6 m
h = 3.03x10^6 m
that's over 3000 km. Did I not use for right equation? Is Eg not 4.5 N/kg?
Also for b), isn't the force of gravity the centripetal...
I've been attempting to solve this problem for three days now. I have thrown away my old attempts (like, scrumpled up into the bin), but my old attempts involved:
Trying to set up simultaeneous equations relating the journeys between EH and FG to find the deceleration, but the reason why this...
So far what we know about the circular motion is that an object moving in a circle experiences a force towards the center of the circle and as a result accelerates towards this center.
But we also know that an object always moves in the direction of resultant force - if two tractors moving at...
ω(10)=(1.3)∗(1.0−e^(−10/22) )= 0.475 rad/s
0.475 rad/s=0 +α(10second)
α=0.0475 rad/s^2
∫ω(t)=Θ =1.3t + 28.6e^(-t/22) | (t=10s, t=0)
total angle by which the wheel rotates over this period of t=10 seconds = 2.55 rad
Θ= 2(pi)(8m)= 1.3t + 28.6e^(-t/22)
0=1.3t + 28.6e^(-t/22) - 2(pi)(8m)
t=34...
Hello, I hope you are all very well.
I am in second year of High School and I have a practical work in physics.
The experiment was to release a long tape with a mass of 40g at the end from a certain height. An instrument would hit the tape 50 times/s and put a mark each time. From that we have...
Well, first a wrote the equation for acceleration in non inertial systems.
##a_I=a_o+\dot \omega \times r+\omega \times (\omega \times r)+2(\omega \times v_{rel}) +a_{rel}##.
Then, ##a_o=0## (because the system doesn't move), ##a_i=0## (because it is measured from the non inertial system)...
Summary: what is the equation for a pumkin's acceleration when the air pressure in not constant.
My Daughter and I are going to Pumpkin chunkin for the first time.
I would like to get two orange shirts and scribble:
what part of this don't you understand:
(DifEq?) acceleration of a pumpkin...
1. For the car to apply brakes, we have ##v^2=2ar⇒a=\frac{v^2}{2r}=μg\;\;[ma=μmg]⇒v=\sqrt{2μgr} ##
2. For the car to go in a circle ##\frac{mv^2}{r}=μmg\Rightarrow v=\sqrt{\mu gr}##.
We find from above that the maximum velocity ##v## possible to avoid a collision is ##\sqrt{2}## times as much...
I tried this problem 3 times. I only have two attempts left.
First time: Centripetal acceleration: 7560 m/s^2
Centripetal Force: 4.7 Newtons
Second time: Centripetal acceleration: 25.032
Centripetal Force: 4.7
Third Time:
Centripetal...
I calculated the average velocity in a previous problem and got 1.146788991m/s over a time of 8.72s. I know I can’t use a_ave=(Vf-Vi)/(Tf-Ti) because I don’t know the final velocity and have no way to find it. Do I multiply average velocity by time?
Summary: Non - ideal pulley question, should be easy but has got me good
Hey guys, looking for some help on this pulley question. It involves torque, and works with Newton's 2nd law in conjunction with a non-ideal pulley.
Text of question:
" When the motor in the figure below lowers the m =...
I'm struggling in the details of this exercise. Let ##S'## be the reference frame where the acceleration of the spaceship is constant, in which case we have ##u'(t')= a' t'## (since we assume no acceleration at the beginning). The rest frame of the rocket ##S## is connected to ##S'## via a...
This question showed up on my grade 12 physics test.
The problem I have with this question is that I did not know the direction that the system would accelerate in, so I just solved as though the mass on the inclined plane would accelerate the system. I expected that if it would accelerate the...
I have some difficulties trying to understand non-inertial frames.
I have problems to notice the acceleration in these cases, from an inertial reference frame and from non inertial refrence frame.
Consider the first case, if I'm on the wedge, I see that the block doesn't move so there's no...
I am trying to solve accelerations of a cart on these different slopes. I don't understand how it is possible without knowing the coefficient of friction, but my teacher says it is (very cryptically I might add). Can anyone help me understand this? Thanks.
I considered the downwards direction and left direction as negative. For ##m_1##, Newton's equations are:
##x) Fr + W_x - T=0##
##y) N - W_y =0##
For ##m_2##:
##y) T - W =0##
Then, if I replace the data, I get ##T=22.2 N## and then ##m_2=2.2 kg##.
With that, for the second question ##m_2=4.4...
What should I do? Because I have two possibilities. I have ##0=5+at## so ##-5/t =a##. But then I can also say that the acceleration is a negative because it is stopping, so I can write it like ##0=5-kt## and then ##5/t =k##
Homework Statement: A student is swinging a ball on a string overhead in a horizontal plane. The string initially has length 𝑙0 and the ball is moving with an initial speed 𝑣0 . The student decides to see how fast they
can spin the ball, so they begin moving it faster and faster with a constant...
Let m be a point test mass. Initially m has velocity Vy in the poisitive y-direction, and zero velocity in the x-direction. At time zero, m is accelerated in the positive x-direction. In the limit as the time goes to infinity, the velecity in the positive x-direction goes to the speed of...
Hello,
Newton's second law, when the mass is constant, tells us that the acceleration ##a=\frac {F}{m}## which produces a simple ODE.
The acceleration is a function that can be constant ##a= constant##, time-dependent ##a(t)##, velocity-dependent ##a(v)##, position dependent ##a(x)##, etc...
This is a homework question from my friend, I found the time but a tough differential equation occurred when I was trying to find accelaration, is there a simple solution for this?
Well, ##r(t)## in ##A## is just a vector ##(0;y)## because is tangent to the trajectory. Then, from the perspective of ##B## the particle moves in an uniform circular motion. Is this right?
The velocity from ##B## must be ##\omega##, right?
And what about acceleration?
I have a series of pulleys where the belt is running around them in a way to describe a sine curve. The pulleys are stationary and the belt is running from left to right. For every particle of the belt I can use standard formula to calculate their normal acceleration, when in contact with the...
Please could I ask for help with the following question:
Part (a) is no problem. Acceleration is the gradient of the graph in regions OA and AB which gives 3 and 0.5
Part (b), I believe, requires me to calculate the greatest and least value of the gradient of the curve in region BC
Part...
Well, what I've done so far is calculating the magnitude of velocity and acceleration replacing ##t=2## in ##\theta (t)## and ##r(t)## so I could get the expressions for ##\dot r##, ##\dot \theta##, ##\ddot r## and ##\ddot \theta##. But that's not my problem... my problem is related to the...
The question comes from a thought experiment of a rocket approaching the Earth accelerating at a constant rate of 1g from say from a hypothetical "earth like planet" near by. . we would be standing on the floor of the upright rocket as it lifts off, if we are standing on a scale, our...
From what I understand,
##a_{r} = v_{tan}^2 /r##
##a_{r} = (r\omega)^2 /r##
##a_{r} = r\omega^2##
##\omega^2 = \frac{a_{r}}{r}##
##\omega^2 = \frac{2+2t}{0.12}##
##\omega = \sqrt{\frac{2+2t}{0.12}}##
##s =\int_{0}^{2} \sqrt{\frac{2+2t}{0.12}}##
After integrating, I still can't seem to get the...
Choice D is obviously wrong therefore leaving us with choices A, B, and C. Can someone explain the relationship of the three variables stated above (mass, volume, and acceleration due to gravity)? Thank you.
2.3.16 A car is traveling at $45 \, km/h$ at time $t=0$ It accelerates at a constant rate of $10 \, km/h\, s$
(a) How fast is the care going at $t=1\, s$?
$$v_t=v_0+at=45+10(1)=55\,\dfrac{km}{h}$$
at $t=2\,s$
$$v_t=v_0+at=45+10(2)=45+20=60\,\dfrac{km}{h}$$
(b) What is its speed at a...
In a circular orbit, the 4-velocity is given by (I have already normalized it)
$$
u^{\mu} = \left(1-\frac{3M}{r}\right)^{-\frac{1}{2}} (1,0,0,\Omega)
$$Now, taking the covariant derivative, the only non vanishing term will be
$$
a^{1} = \Gamma^{1}_{00}u^{0}u^{0} + \Gamma^{1}_{33}u^{3}u^{3}
$$...
eb2
A car, start from rest, accelerates in straight line at a constant rate of $2.0 m/s^2$
How far will the car travel in 10 seconds
use
$d=vt+\dfrac{1}{2}at^2\quad v=0\quad a=2\,m/s^2\quad t=10$
then
$$d=(0)(10\, s)...