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.
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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etotheipi said:
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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etotheipi said:
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.

aspodkfpo said:
Nope, haven't covered much calculus.

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

aspodkfpo said:
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

aspodkfpo said:
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}##.

1. What is the formula for calculating average speed with constant acceleration over a period of time?

The formula for calculating average speed with constant acceleration over a period of time is: average speed = (final velocity + initial velocity) / 2.

2. How is acceleration related to average speed?

Acceleration is the rate of change of velocity, and it directly affects the average speed of an object. The greater the acceleration, the greater the change in velocity and therefore the greater the average speed.

3. Can the average speed with constant acceleration be negative?

Yes, the average speed with constant acceleration can be negative. This would occur if the object is decelerating, or slowing down, over the period of time.

4. What units are used for average speed with constant acceleration?

The units for average speed with constant acceleration are typically meters per second (m/s) or kilometers per hour (km/h), depending on the units used for time and distance.

5. How does the duration of time affect the average speed with constant acceleration?

The longer the duration of time, the greater the distance the object will cover and the higher the average speed will be. This is because the object has more time to accelerate and cover a greater distance.

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