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Video: Acceleration of Moving Objects

Ed Burdette

In this video we learn how to calculate the motion of an accelerating object that already has a nonzero initial velocity, and derive an equation relating the initial and final velocities of an object undergoing acceleration.

13:37

Video Transcript

In this video, we’re going to learn about the acceleration of moving objects. We’ll learn what this means, how to calculate it, and how to solve problems involving moving objects that accelerate.

To start out, imagine that we’re watching a nature documentary that’s following a snake around at night as it moves through the woods looking for an evening meal. After searching for some time, the snake comes across a rabbit nibbling away at some grass and unaware of its approaching predator. The snake seized its chance and begins to slither toward the rabbit more quickly. Moving along at one meter per second, the snake gets within range and decides it’s time to strike. Suddenly springing forward with an amazing acceleration of 279 meters per second squared, the snake accelerates this way for one quarter of a second, after which it’s one-half meter away from its prey. If it maintains that speed over this distance, then the question is, how much time does the rabbit have to leap to safety?

To answer this question, we want to understand the acceleration of moving objects. To get to the acceleration of objects that are already moving, let’s talk a bit about the term acceleration. As an equation, we can write that acceleration, 𝑎, is equal to a change in velocity, Δ𝑣, divided by the time it takes for that change to occur, Δ𝑡. We can expand this expression to say that acceleration is equal to final velocity minus initial velocity divided by final time minus initial time. And we know that the units of acceleration are meters per second per second, or meters per second squared. We can take this expression for acceleration, that it’s equal to 𝑣 sub 𝑓 minus 𝑣 sub 𝑖 over 𝑡 sub 𝑓 minus 𝑡 sub 𝑖, and rearrange it slightly.

We can write that acceleration equals final velocity minus initial velocity over Δ𝑡. And if we multiply both sides of this equation by Δ𝑡, we see it cancels on the right-hand side. If we then take the step of adding the initial velocity, 𝑣 sub 𝑖, to both sides of the equation, we see that that also cancels on the right-hand side of our equation. This leaves us with an expression reading: final velocity is equal to initial velocity plus acceleration times time. Notice a couple things about this equation. First, it assumes that the acceleration, 𝑎, is constant over the time it acts, Δ𝑡. And second, the units of acceleration, meters per second squared, when we multiply them by the units of time, seconds, become units of meters per second, consistent with the units in the rest of our expression.

This expression is useful to us in analyzing the acceleration of moving objects because it doesn’t assume that the object’s initial velocity is zero. It could have any value. And given that and acceleration and a change in time, we’re still able to solve for the final velocity achieved. Let’s get some practice using this expression in studying moving object acceleration with a few examples.

A rocket sled accelerates from rest to a top speed of 264 meters per second in 4.38 seconds and is brought jarringly back to rest in 1.06 seconds. Calculate acceleration in the direction of the rocket sled’s motion, giving your answer as a multiple of 𝑔, where 𝑔 is assumed to equal 9.8 meters per second squared. Calculate acceleration in the direction opposite to the rocket sled’s motion, giving your answer as a multiple of 𝑔, where 𝑔 is assumed to equal 9.8 meters per second squared.

As the rocket sled in this example moves in a particular direction, it accelerates forward to a top speed of 264 meters per second over a time of 4.38 seconds. Then it accelerates opposite that direction over a time of 1.06 seconds, coming back to rest. We want to solve for acceleration both in the forward direction of the sled’s motion and in the opposite direction of the sled’s motion. We’ll give our answers in terms of 𝑔, the acceleration due to gravity, which we’ll assume is exactly 9.8 meters per second squared. We can call 𝑎 sub 𝑓 the acceleration of the rocket sled in the forward direction and 𝑎 sub 𝑟 the acceleration of the rocket sled in the reverse direction.

Let’s start by drawing a sketch of the motion of the rocket sled. We begin our scenario with a rocket sled at rest on a set of rails. Its engine is powering up and it’s just about to accelerate to the right in forward motion. The sled begins to accelerate forward and does so until it reaches a maximum speed of 264 meters per second over a time of 4.38 seconds. The acceleration it experiences over this time, we’ve called 𝑎 sub 𝑓. After achieving this maximum speed, the rocket sled puts on the brakes and begins to accelerate in the direction opposite of motion. It’s slowing down, eventually coming to a stop. It does this over a time interval we can call 𝑡 sub two of 1.06 seconds. As we solve for 𝑎 sub 𝑓 and 𝑎 sub 𝑟, we want to report those results in terms of the unit of acceleration due to gravity, 𝑔, which is exactly 9.8 meters per second squared.

To begin solving for these terms, we can recall that acceleration, 𝑎, is equal to final velocity minus initial velocity divided by the time change it takes for that velocity change to occur. When we write this expression for our given information, we see that 𝑎 sub 𝑓, the acceleration when the rocket sled is in forward motion, is equal to the final velocity it achieves, we’ve called it 𝑣 sub max, minus its initial velocity, which is zero since the rocket sled started at rest, divided by the time it took for that velocity change to occur. We’ve called it 𝑡 sub one. If we calculated 𝑎 sub 𝑓 as it’s written, we would get a result in meters per second squared. But we want to express our answer in terms of 𝑔. So to do that, we’ll add that factor, the acceleration due to gravity, into the denominator of our equation.

When we plug in 𝑣 sub max, 𝑡 sub 𝑖, and 𝑔 — notice that when we calculate this value, we’ll have an answer that has no units. The units in all these terms cancel one another out. When we calculate this fraction and round it to two significant figures, we find that 𝑎 sub 𝑓 is 6.2 times the acceleration due to gravity, 𝑔. That’s the acceleration the rocket sled experiences when it’s in forward acceleration.

Now we move on to solving for 𝑎 sub 𝑟, the sled’s acceleration in the reverse direction of its motion. That is, as it comes to a stop. When we apply the mathematical relationship for acceleration to 𝑎 sub 𝑟, we find that its final velocity, zero meters per second, minus its initial velocity, 264 meters per second, is then divided by what we’ve called 𝑡 sub two, the time it takes for the sled to come to a stop. Once again, because we want our answer to be in terms of 𝑔, we add that factor to the denominator of our expression.

When we plug in for 𝑡 sub two and 𝑔, we see again that the units in this expression cancel out. And we also observe that because we’re giving our final answer in terms of 𝑔, we can solve for the magnitude of 𝑎 sub 𝑟 and change the negative sign in the numerator to positive. Calculating this fraction, we find, to two significant figures, it’s 25. Which means that 𝑎 sub 𝑟 is 25 times 𝑔. This is the acceleration of the sled as it slows down.

Now let’s look at an exercise where the initial speed of our object is not zero. That is, it’s already moving.

A particle moves in a straight line with an initial velocity of 30.0 meters per second. It accelerates for 5.00 seconds at a constant 30.0 meters per second squared in the direction of its initial velocity. What is the magnitude of the particle’s displacement after the acceleration? What is the magnitude of the particle’s velocity after the acceleration?

In this two-part problem, we want to solve for the particle’s displacement after it undergoes its acceleration, which we can call 𝑑. And we also wanna solve for its velocity after this acceleration occurs. We can call this 𝑣 sub 𝑓. In terms of the particle’s motion, we’re told it starts out with an initial velocity of 30.0 meters per second. And then, for 5.00 seconds, it accelerates at 30.0 meters per second squared. We can come to a better understanding of the particle’s motion by drawing a graph of its velocity versus time.

Here we have a chart of the particle’s velocity versus time, where time, in units of seconds, is on the horizontal axis and velocity, in units of meters per second, is on the vertical axis. We’re told that our particle begins with a velocity, we’ve called it 𝑣 sub 𝑖, of 30.0 meters per second. So we mark that value in as our initial velocity on our chart. We’re told further that our particle that accelerates at an acceleration of 30.0 meters per second squared for a time, we’ve called 𝑡 sub 𝑎, of 5.00 seconds.

When we write out our acceleration of 30.0 meters per second squared, we understand that that means 30.0 meters per second every second. In other words, for every one second of time passed, we add 30.0 meters per second to our velocity. So if the next tick mark on our vertical axis is 60 meters per second, then we achieve that velocity at a time of one second. And under this continuing acceleration of 30.0 meters per second, we continue to add 30.0 meters per second to our velocity every second. By connecting these dots, we now have our velocity versus time curve for the particle. And we see that under these conditions, the maximum velocity we achieve is 180 meters per second. That’s our result for 𝑣 sub 𝑓, the final velocity achieved.

In our case, we’ve solved for this result graphically. That’s another way of finding the same result.

Now we want to solve for the total displacement of the particle. We’ve called it 𝑑. We can recall that an object’s displacement is equal to the integral of its velocity with respect to time. And our graph is a velocity versus time curve. This means that if we solve for the area under our velocity versus time curve, then that total area result will give us our displacement 𝑑. To make calculating this total area easier, we can divide the area up into two pieces, a rectangular piece as well as a triangular piece above that.

We can now write that our displacement 𝑑 is equal to the area of the rectangle plus the area of the triangle in units of meters. We can write that the area of our rectangle is equal to its height, 30.0 meters per second, times it’s width, 5.00 seconds. Notice the units of seconds cancel, and we’re left with units of meters. Then we add that area to the area of our triangle which is one-half its base, 5.00 seconds, times its height, 180 minus 30, or one 150 meters per second. Again, the units of seconds cancels out, and we’re left with units of meters. When we add these two values together, we find a result of 525 meters. That’s the particle’s total displacement after these five seconds of acceleration and starting at our initial velocity 𝑣 sub 𝑖.

Let’s summarize what we’ve learned about the acceleration of moving objects. We’ve seen that when a moving object accelerates, we need to be careful to account for its initial motion in order to figure out, or solve, for where it ends up. We’ve seen that by rearranging an expression for acceleration, we can find an equation which, when acceleration is constant, will help us solve for final velocity, given an initially moving object.

And we haven’t forgotten about our rabbit and our snake. We recall the snake was in the middle of trying to catch a midnight snack. Given the snake’s initial velocity, its acceleration, the time it sustained that acceleration, and the distance it had to get to the rabbit, we find that the time the rabbit has to get away is a mere 7.1 milliseconds. It will have to act fast.

By understanding the acceleration of objects already in motion, we broaden the scope of physical scenarios we’re able to understand and solve for.