Question Video: Acceleration over a Distance | Nagwa Question Video: Acceleration over a Distance | Nagwa

# Question Video: Acceleration over a Distance Physics

An object has an initial velocity of 6 m/s and accelerates in the direction of its velocity along a 16 m long straight line at a rate of 2 m/s². What final velocity does the object have?

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### Video Transcript

An object has an initial velocity of six meters per second and accelerates in the direction of its velocity along a 16-meter-long straight line at a rate of two meters per second squared. What final velocity does the object have?

Okay, so in this question, there’s a lot of information. So let’s underline all the important bits. So we know that we’ve got an object which has an initial velocity of six meters per second. And it accelerates in the direction of its velocity along a 16-meter-long straight line. The rate of its acceleration is two meters per second squared. We need to find the final velocity of the object. Let’s call this quantity 𝑣, the final velocity. We can also label all of the other information we’ve got in a very concise form, just below the 𝑣.

For example, we know the initial velocity of the object. Let’s call this 𝑢. 𝑢 happens to be six meters per second. But why are we using the letter 𝑢? Well, we’ve said that the final velocity is 𝑣. And we need another letter to describe the initial velocity. Since 𝑢 comes before 𝑣 in the alphabet and the initial velocity comes before the final velocity in time, then 𝑢 is as good a choice as any other letter. Anyway, let’s also label the acceleration of the object. We’ll call this 𝑎. And it’s two meters per second squared. And the distance that the object travels, 𝑠, is 16 meters. We’re using the letter 𝑠, by the way, because this distance is also a displacement.

The object is travelling in a straight line between its start point and its finish point. Now the shortest distance between two points is also a straight line. And a displacement is defined as the shortest distance between the start point and the finish point of any path taken by an object. That shortest distance, as we’ve already mentioned, happens to be a straight line. Hence, the distance that this object is travelling is also a displacement. Conventionally, we label displacement with the letter 𝑠. Therefore, we’ve used the letter 𝑠 here.

Now, let’s very quickly discuss the fact that the acceleration is in the direction of the velocity of the object. Let’s say we’ve got this object here, the blue blob. Initially, it’s moving at six meters per second to the right, for example. Then it accelerates at two meters per second squared. And eventually, it reaches its finishing point. This finishing point is 16 meters away from its start point. What we need to do is to find the final velocity of this object 𝑣. That is, the velocity it’s travelling at when it’s travelled the 16 meters. Now, the acceleration being in the direction of the object’s velocity means that in this case, the acceleration is also to the right. The object is speeding up. Therefore, the acceleration is positive. If the acceleration was to the left or the object was slowing down, then it will be negative. But because it’s in the direction of its velocity, we’re gaining speed. And so the acceleration is positive.

Okay, now what we can do is to replace all of the quantities in this diagram with their symbols, which makes it a lot less messy. Then, we can actually go on to solving this problem. We need a relationship that links together the final velocity, 𝑣; the initial velocity, 𝑢; the acceleration, 𝑎; and the distance, 𝑠. But what is this relationship? Well, let’s start by saying that there is a reason why we labeled these quantities as 𝑣, 𝑢, 𝑎, and 𝑠. We did this so that we can use the SUVAT equations. The SUVAT equations are equations of motion. That is, they describe how things move when the acceleration is constant. And in this scenario, that’s exactly what we’ve got. We’ve got a constant acceleration of two meters per second squared. Therefore, we can use the SUVAT equations.

But why are they called SUVAT equations? Well, it’s because they deal with quantities, such as the distance, 𝑠; the initial velocity, 𝑢; the final velocity, 𝑣; the acceleration, 𝑎; and the time taken, 𝑡. Now in this case, we’re not dealing with the time taken, 𝑡, neither are we given it nor are we trying to find it out. So we need to pick the correct SUVAT equation that only deals with these four quantities: 𝑣, 𝑢, 𝑎, and 𝑠. Any of the other SUVAT equations are not useful to us in this situation because they have the time 𝑡 in them.

So let’s write down the SUVAT equation that deals with 𝑣, 𝑢, 𝑎, and 𝑠 only. That equation is the following: 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. The final velocity squared is equal to the initial velocity squared plus two times the acceleration times the distance. And this is the correct equation because it’s got one value that we don’t know and three that we do. Which means that we can start plugging in values. But before we do, let’s discuss the units of each one of these quantities.

It’s important that we try to work in the standard units of these quantities. 𝑢 and 𝑣, the two velocities, should be in meters per second. 𝑎, the acceleration, should be in meters per second squared. And 𝑠, the distance, should be in meters. And we could see on the left here, that’s exactly what we’ve got. The three quantities that we do know are in the correct standard units. Which means that when we calculate the final velocity 𝑣, that will also end up being in meters per second. So without further ado, let’s start calculating.

We need to start by rearranging this equation very slightly so that we can solve for the final velocity 𝑣. We can do this by taking the square root of both sides of the equation. On the left-hand side, the square root cancels with the square. So we’re left with 𝑣 is equal to the square root of 𝑢 squared plus two 𝑎𝑠. Now, we can put our values in. 𝑣 is equal to the square root of 𝑢 squared, which is six squared plus two times 𝑎, which is two times 𝑠, which is 16. And this simplifies slightly to 𝑣 is equal to the square root of 100. Taking the square root of 100, we find that the value of 𝑣 could be either positive 10 or negative 10. We’ve shown this in our working out. But we need to figure out if both of these answers are correct.

So let’s start with positive 10 first. We know that we’ve got an object here, starting at six meters per second, moving to the right. And it accelerates in the direction of its velocity. That means it’s getting faster. Its speed is increasing. So after it’s travelled this distance of 16 meters, does it make sense for the object to now be travelling at 10 meters per second? Well, yes, it does. The object was accelerating. And 10 meters per second is faster than six meters per second. Therefore, the positive value checks out.

Let’s now look at the negative value. Once again, the object is starting here, moving towards the right at six meters per second. It then accelerates in the direction of its motion, so speeds up. And then it’s travelling at negative 10 meters per second. What? It’s going to the left! Cause that’s what negative means, right? It means it’s travelling in the opposite direction. So how can an object be accelerating in its direction of motion, so gaining speed, and still end up going the other way. That doesn’t make sense. And so the negative answer doesn’t make sense. It’s not a physical answer. We can ignore it. Therefore, we can go on to writing down our final answer.

The final velocity is 10. And we don’t need to use the positive sign, by the way, because now we know that that’s the only possible answer. And we need to put the units in, of meters per second, as we discussed earlier. So the object that starts out with a velocity of six meters per second and accelerates in the direction of its velocity along a 16-meter-long straight line at a rate of two meters per second squared, has a final velocity of 10 meters per second.

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