Acceleration/Velocity Relationship

[tascja]

i was wondering if someone could explain the relationship between distance/velocity/acceleration? I don't have a clear understanding how they are related.. especially when they are plotted on a graph.

Ill give you a specific example:
so there was a velocity-time graph where the average acceleration (slope) over the entire graph was 0.5m/s^2. The instantaneous acceleration at t=16s was -1m/s^2. And it asked if the average acceleration was greater than the instantaneous or was the instantaneous greater than the average?
My thoughts were that since the instantaneous acceleration was greater in magnitude it was greater... this was wrong :( but i don't understand how or why??

 

[LowlyPion]

Velocity is the Δd/Δt. It is the time rate of change of displacement.

Acceleration the is the Δv/Δt = Δ (Δd/Δt) /Δt or Δ²d/Δt²

Both of these are vectors - and direction matters just like the displacement vector d that they are expressions of. You should view them as a continuum.

Hence -1 is greater than -2 but less than .5.

 

[nlightner]

The relationship between distance, velocity, and acceleration is an important concept in physics and can be explained through the equations of motion. Let's break down the relationship between these three quantities.

Distance is the total length of the path traveled by an object. It is a scalar quantity, meaning it has only magnitude and no direction.

Velocity is the rate of change of an object's position over time. It is a vector quantity, meaning it has both magnitude and direction. Velocity is calculated by dividing the change in distance by the change in time, and is represented by the slope of a distance-time graph.

Acceleration is the rate of change of an object's velocity over time. It is also a vector quantity and is calculated by dividing the change in velocity by the change in time. On a velocity-time graph, acceleration is represented by the slope of the velocity-time curve.

Now, let's look at the specific example you provided. The average acceleration over the entire graph was 0.5m/s^2. This means that on average, the velocity of the object was increasing by 0.5m/s every second. However, at t=16s, the instantaneous acceleration was -1m/s^2. This means that at that specific moment, the velocity of the object was decreasing by 1m/s every second.

So, to answer the question of whether the average acceleration was greater than the instantaneous or vice versa, we need to consider the magnitudes of these values. In this case, the instantaneous acceleration (-1m/s^2) is greater in magnitude than the average acceleration (0.5m/s^2). This is because the object experienced a larger change in velocity over a shorter period of time at t=16s.

I hope this explanation helps clarify the relationship between distance, velocity, and acceleration and how they are related on a graph. It is important to remember that these quantities are all connected and can help us understand the motion of objects in the physical world.