The acceleration at instantaneous maximum velocity is zero, as the velocity is not changing at that moment.
"Acceleration" means change of velocity. If velocity is constant, then acceleration is zero.
No, if velocity is zero and acceleration is less than zero, it means that the object is slowing down. Speed is the magnitude of velocity, so if acceleration is negative, the speed decreases.
Since the derivative of velocity is acceleration, the answer would be technically 'no'. Here is why: v = 0 v' = 0 = a Or in variable form... v(x) = x v(0) = 0 v'(0) = 0 = a You can "trick" the derivative into saying that v'(x) = 1 = a (since the derivative of x = 1) and then stating v'(0) = 1 = a... but that is not entirely correct. Acceleration is a change over time and is measured at more then one point (i.e. the acceleration of this body of matter is y from time 1 to 5) unless using derivatives to form the equation of the acceleration line/curve. If an object has a constant acceleration of 1, then the velocity is constantly increasing over that time. Using the equation discussed above and looking at acceleration over time, at 0 seconds, acceleration is 0 and so is velocity, but from 0-1 seconds acceleration is 1 and velocity is 1 as well. 0-2 seconds, acceleration is 1, but velocity would be 2 (at the end of 2 seconds).
The instantaneous acceleration of the particle is equal to 0 when the velocity of the particle is at a maximum or minimum. This occurs at the points on the graph where the slope of the velocity-time graph is horizontal or the velocity reaches a peak or trough.
When a pendulum reaches its maximum elongation the velocity is zero and the acceleration is maximum
0 velocity
Yes, the velocity of the chair changes as a simple harmonic motion (SHM) and varies from 0 to maximum back to 0 back to ... The acceleration of the chair also relates to SHM in that it varies inversely with the velocity - high velocity = 0 acceleration, 0 velocity = maximum acceleration. If you chart the speed of the shadow of the chair you will see a smooth waveform - likely a sine wave.
The acceleration at instantaneous maximum velocity is zero, as the velocity is not changing at that moment.
I'm trying to give you a simple example. Which hope will be able to make you understand that thing. We read physics in our language. So I can make some mistakes to write that in English... But I hope I won't be mistaken. By the way here it is... Suppose, There is a simple oscillator what is moving under the angles of 4 degrees. When it is in his height position then we know that it stops for a moment and then comes back to the equilibrium position. So in the height position it's velocity is 0 m/s. (cause it stops.) Suppose it's 1st velocity or the velocity in the equilibrium position is "v" and in that position the last velocity is 0. so we can calculate it's acceleration like that a = (last velocity-1st velocity)/time so a = (0-u)/t = - u/t, what is the maximum acceleration of a this body, (-) sign means the acceleration is gonna go down. But when it is coming back to the equilibrium position, it's velocity goes up.. And in the equilibrium position it's velocity is maximum. Then the velocity decreases again. so in the equilibrium position the velocity is maximum, in the equilibrium position which is "v". so it's acceleration will be a= v-v/t = 0/t = 0, that means the acceleration is zero when the velocity is maximum... That means the acceleration of a body can be zero when the velocity is maximum and the velocity can be zero when the acceleration is maximum. [Note: Always remember, to start to calculate from the equilibrium position. Because the oscillation starts from the equilibrium position. And what I said is the most simple statement. It can also be described by the equations of the a simple harmonic oscillations] - by JAS
The condition for maximum velocity is acceleration equals zero; dv/dt = a= o.
"Acceleration" means change of velocity. If velocity is constant, then acceleration is zero.
the acceleration is equal to energy that release by the friction that came be electic that travel form somewhere.It proves that maximum acceleration rate.The easy explainationof that is Energy and Velocity are equal to maximum of acceleration
No, if velocity is zero and acceleration is less than zero, it means that the object is slowing down. Speed is the magnitude of velocity, so if acceleration is negative, the speed decreases.
Since the derivative of velocity is acceleration, the answer would be technically 'no'. Here is why: v = 0 v' = 0 = a Or in variable form... v(x) = x v(0) = 0 v'(0) = 0 = a You can "trick" the derivative into saying that v'(x) = 1 = a (since the derivative of x = 1) and then stating v'(0) = 1 = a... but that is not entirely correct. Acceleration is a change over time and is measured at more then one point (i.e. the acceleration of this body of matter is y from time 1 to 5) unless using derivatives to form the equation of the acceleration line/curve. If an object has a constant acceleration of 1, then the velocity is constantly increasing over that time. Using the equation discussed above and looking at acceleration over time, at 0 seconds, acceleration is 0 and so is velocity, but from 0-1 seconds acceleration is 1 and velocity is 1 as well. 0-2 seconds, acceleration is 1, but velocity would be 2 (at the end of 2 seconds).
Not at all. Zero acceleration just means that the velocity is not changing ...the motion is in a straight line at a consgtant speed.
The instantaneous acceleration of the particle is equal to 0 when the velocity of the particle is at a maximum or minimum. This occurs at the points on the graph where the slope of the velocity-time graph is horizontal or the velocity reaches a peak or trough.