Lesson Video: Average and Instantaneous Acceleration

In this video we learn about average and instantaneous acceleration; we learn how to calculate them and how they relate to an object’s position and velocity.

09:08

Video Transcript

In this video, we’re going to learn about average and instantaneous acceleration. What these terms are, how they relate to velocity and position, and how to use them practically in solving problems.

As we start off, imagine you’re at the racetrack watching your favorite driver compete in a championship race. After several hours of driving, the cars all come down to their final lap around the racecourse. Unfortunately, your favorite driver is not in front. And you begin to wonder, if all the other cars maintain their current speed, then would it even be possible for your driver to accelerate past them and cross the finish line first? This is a question we’ll need to understand acceleration to be able to answer.

Acceleration is the time rate of change of velocity. This means that if we wanted to solve for an object’s average acceleration, then over some time interval, starting with 𝑡 initial and ending with 𝑡 final, we would subtract the object’s velocity at the initial time from the object’s velocity at the final time and divide that by the difference in time total.

If we were to extend this notion to calculate an instantaneous acceleration, that is acceleration at a specific point in time, then we would deliberately shrink the gap between the final and the initial time, smaller and smaller until that time differential approached zero. Now, we know that instantaneous acceleration is equal to a time derivative of velocity. And we may recall that instantaneous velocity is given as a time derivative of position.

So, we now have three terms, position, velocity, and acceleration, that all are connected to one another. We could almost think of it like a ladder, where position is on the top rung, velocity is on the next rung, and acceleration is below that. To get from position to velocity, we would take a derivative with respect to time. And likewise, to get from velocity down to acceleration. And if we wanted to move up the rungs of the ladder, say from acceleration to velocity, then we take an integral of acceleration with respect to time. And similarly, to get from velocity to position.

As a side note, these three terms aren’t the only terms that describe motion. If we were to take the time derivative of acceleration, we arrive at a lesser-known term known as jerk. If we were riding in a car whose acceleration was changing, then this term, jerk, might be a good description for how it would feel. That said, position, velocity, and acceleration cover most of the motion that we’re interested in. Now, that we know how to calculate average and instantaneous acceleration, let’s get some practice using these concepts.

A racehorse passing through a starting gate accelerates from rest to a velocity of 15.0 meters per second due west, in a time interval of 1.80 seconds. Find the racehorse’s average acceleration. Assume that east corresponds to positive displacement.

We want to solve for the average acceleration of the racehorse over this 1.80-second time interval. We can call that time interval Δ𝑡. And we can call the final velocity of the racehorse, 15.0 meters per second, 𝑣 sub 𝑓. If we call 𝑎 sub avg the average acceleration of the racehorse, we can write that that’s equal to the racehorse’s final velocity minus its initial velocity over Δ𝑡, the time it took to make that change in velocity.

Since the horse was passing through the starting gate, we can assume that 𝑣 sub 𝑖 is zero. And since we’re given 𝑣 sub 𝑓 and Δ𝑡 in the problem statement, we’re almost ready to solve for 𝑎 sub avg. The one point we want to keep in mind is that this is a vector. And therefore, it could have positive or negative value.

We’re told that our velocity, 𝑣 sub 𝑓, is in the west direction but that motion to the east is motion that we can consider in the positive direction. So, this means we can insert a minus sign in front of our final velocity of the racehorse. After all, it’s moving in the negative direction, by our convention. So, the acceleration on average is negative 15.0 meters per second divided by 1.80 seconds, or negative 8.33 meters per second squared. That’s the average acceleration of the racehorse.

Now, let’s look at an exercise that lets us solve for instantaneous acceleration, as well as instantaneous velocity.

The position of a particle along the 𝑥-axis varies with time according to the equation 𝑥 as a function of 𝑡 equals 1.5 minus 3.3𝑡 squared meters. What is the velocity of the particle at 𝑡 equals 2.7 seconds? What is the velocity of the particle at 𝑡 equals 4.3 seconds? What is the acceleration of the particle at 𝑡 equals 2.7 seconds? What is the acceleration of the particle at 𝑡 equals 4.3 seconds?

So, given an equation describing the position of a particle as a function of time, we want to solve for instantaneous velocity and instantaneous acceleration at two different time values, 𝑡 equals 2.7 seconds and 𝑡 equals 4.3 seconds. We can name the values we want to solve for 𝑣 of 2.7 seconds, 𝑣 of 4.3 seconds, 𝑎 of 2.7 seconds, and 𝑎 of 4.3 seconds.

Given this information, we can recall that there is a relationship between position, velocity, and acceleration. Instantaneous velocity is equal to the time derivative of position. And instantaneous acceleration is equal to the time derivative of velocity, which is also equal to the second time derivative of position. Since we’re given the position of our particle with respect to time, let’s differentiate it twice to solve for instantaneous velocity and instantaneous acceleration.

Instantaneous velocity as a function of time is equal to the time derivative of 1.5 minus 3.3𝑡 squared meters. Calculating this derivative, we find it’s equal to negative 6.6𝑡 meters per second. This tells us that, to solve for velocity when 𝑡 equals 2.7 seconds and when 𝑡 equals 4.3 seconds, we only need to plug in those time values into this general expression. When we do, to two significant figures, the velocity at 2.7 seconds is negative 18 meters per second and the velocity at 4.3 seconds is negative 28 meters per second.

Now, we move on to solving for acceleration at these two particular time values. Recalling that acceleration is equal to the time derivative of velocity, we can write that acceleration as a function of time is equal to the time derivative of negative 6.6𝑡 meters per second. We find that this derivative is a constant, negative 6.6 meters per second squared. This means that for any time value, that will be our acceleration. Therefore, the acceleration at 2.7 seconds equals the acceleration at 4.3 seconds, which is equal to negative 6.6 meters per second squared. That’s the acceleration at these two time values.

Let’s take a minute to summarize what we’ve learned about average and instantaneous acceleration. Acceleration is equal to the time rate of change of velocity. That is, it’s a vector. Average acceleration is equal to the velocity of an object at some final time minus the velocity of that same object at an initial time all divided by that time interval, 𝑡 final minus 𝑡 initial.

Instantaneous acceleration is equal to the time derivative of velocity, 𝑑𝑣 𝑑𝑡. These two equations are connected. As the time interval between 𝑡 final and 𝑡 initial gets smaller and smaller approaching zero, average acceleration approaches instantaneous acceleration.

And acceleration is connected with velocity and position through derivatives with respect to time. Given one of these three quantities, we can solve for the others either by integrating or differentiating with respect to 𝑡. Typically, when we know an object’s position, velocity, and acceleration as a function of time, we’re able to understand its motion completely.

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