# Average speed with constant acceleration over a period of time

• aspodkfpo
In summary, the average acceleration of an object undergoing SHM can be calculated by taking the difference of the cosine of the angular frequency at the start and end times, multiplying it by the amplitude and angular frequency, and dividing it by the time interval. This value may differ from the average acceleration calculated using the initial and final acceleration values.f

#### aspodkfpo

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Say that we have an instance where something falls down from a certain height with constant acceleration g. We know that the average speed with regards to the time period is less than (u+v)/2 since we spend less time at the higher speeds.

How do we actually calculate the average speed over a time period,t?

In a general case (but with the constraint that position increases monotonically) the average speed will be$$\langle v \rangle = \frac{1}{T} \int_0^T v(t) dt = \frac{1}{T} = \frac{1}{T}\left[ x(t) \right]_0^T = \frac{\Delta x}{T}$$i.e. it is the distance traveled divided by the total time. But as a matter of fact, in your example the speed even increases uniformly, so why do you think the average speed will not be ##\frac{1}{2}(u+v)##? Less time will not be spend on higher speeds, the profile will be uniform.

You can try calculating it with the other way and see that you get the same answer.

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In a general case (but with the constraint that position increases monotonically) the average speed will be$$\langle v \rangle = \frac{1}{T} \int_0^T v(t) dt = \frac{1}{T} = \frac{1}{T}\left[ x(t) \right]_0^T = \frac{\Delta x}{T}$$i.e. it is the distance traveled divided by the total time. But as a matter of fact, in your example the speed even increases uniformly, so why do you think the average speed will not be ##\frac{1}{2}(u+v)##? Less time will not be spend on higher speeds, the profile will be uniform.

You can try calculating it with the other way and see that you get the same answer.

Oops, I was thinking about a different scenario with a changing force.

Say that I have a spring and an object attached to it and it is pushed inwards by a distance x. The force varies with distance. While the average acceleration of the object over this distance is (0+a)/2, this is not true for the average acceleration over this period of time.

Was wondering how would we get the average acceleration over this period of time, and how this value would compare to (ai + af)/2.
How would the time for it to travel a distance with a changing force compare to the time for constant acceleration of a/2?

Right, so the object is at the end of the spring, you compress it by an amount ##x_0## and then release the object? Between what times do you want the average acceleration; say compressed to zero-extension state (i.e. one quarter cycle)?

You could start off with Newton II,$$-kx = m\ddot{x}$$That gives you the equation of motion,$$x = x_0\cos{\omega t}$$ with ##\omega = \sqrt{\frac{k}{m}}##. Differentiate twice for ##a_x(t)##,$$a_x(t) = -\omega^2 x_0\cos{\omega t}$$The average acceleration during the quarter cycle will be$$\langle a_x(t) \rangle = \frac{4}{T}\int_0^\frac{T}{4} -\omega^2 x_0 \cos{(\omega t)} dt$$Can you evaluate this integral?

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Right, so the object is at the end of the spring, you compress it by an amount ##x_0## and then release the object? Between what times do you want the average acceleration; say compressed to zero-extension state (i.e. one quarter cycle)?

You could start off with Newton II,$$-kx = m\ddot{x}$$That gives you the equation of motion,$$x = x_0\cos{\omega t}$$ with ##\omega = \sqrt{\frac{k}{m}}##. Differentiate twice for ##a_x(t)##,$$a_x(t) = -\omega^2 x_0\cos{\omega t}$$The average acceleration during the quarter cycle will be$$\langle a_x(t) \rangle = \frac{4}{T}\int_0^\frac{T}{4} -\omega^2 x_0 \cos{\omega t}$$Can you evaluate this integral?

Nope, haven't covered much calculus.

Nope, haven't covered much calculus.

Aren't integrals generally in the form of bigf f(x)dx? where is the dx gone here.

Aren't integrals generally in the form of bigf f(x)dx? where is the dx gone here.

Sorry, it was a typo; I've fixed it now Was wondering how would we get the average acceleration over this period of time, and how this value would compare to (ai + af)/2.
If this is SHM, the displacement at time t is ##A\sin(\omega t)## and the speed is ##A\omega\cos(\omega t)##. So from ##t_1## to ##t_2## the average acceleration is ##\frac{A\omega(\cos(\omega t_2)-\cos(\omega t_1))}{t_2-t_1}##.