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Video: Acceleration over a Distance

Parth Gharfalkar

An object has an initial velocity that increases to 14 m/s as the object accelerates in the direction of its velocity. The object accelerates along a 17.1 m long straight line at a rate of 5 m/s². What is the object’s initial velocity?

04:21

Video Transcript

An object has an initial velocity that increases to 14 meters per second as the object accelerates in the direction of its velocity. The object accelerates along a 17.1-meter-long straight line at a rate of five meters per second squared. What is the object’s initial velocity?

Alright, so in this question, we’re trying to find out the object’s initial velocity. We’re told that the object starts with this initial velocity. And then, its velocity increases to 14 meters per second. The way it does this is by accelerating in the direction of its velocity. And this acceleration is five meters per second squared. We’re also told that the object is moving along a 17.1-meter-long straight line. So let’s label all of this information with symbols.

Firstly, we’re trying to find out the object’s initial velocity. Let’s call this 𝑢. And we don’t know what that is, so a question mark. Secondly, we know the final velocity of the object. Let’s call this 𝑣. And we know that that’s 14 meters per second. Thirdly, we also know the acceleration of the object. We’re gonna call this 𝑎. And we know that 𝑎 is equal to — now, we’re told that the object accelerates in the direction of its velocity. So if the object is moving towards the right, then it also accelerates towards the right. Of course, it doesn’t say anywhere that it’s moving towards the right. But we can choose which direction the object is moving. The point is that the acceleration is in the same direction as its velocity. So the value of the acceleration is positive. And this is positive five meters per second squared.

Very finally, we know that the distance that the object is moving, we’ll call this 𝑠, is 17.1 meters. And this distance is a straight line. The fact that the object is moving along a straight line is important because it means that it can have a constant acceleration.

Remember, acceleration is the rate of change of velocity. And velocity is speed in a given direction. So acceleration doesn’t just mean when you speed up or slow down. You can also accelerate if you change direction. However, in this case, we’ve got an acceleration which is five meters per second squared. So it’s constant. And because the object is moving in a straight line, it’s not changing direction either. So the acceleration is overall a constant.

This becomes important because at this point, we can use a set of equations known as the SUVAT equations. The SUVAT equations are a set of equations of motion. That is, they describe how objects move, provided that the acceleration of the object is constant. And as we’ve just seen, it is. Which means that we can use the SUVAT equations in this scenario. Now, the reason that SUVAT equations are called SUVAT equations is because they deal with quantities, such as the distance travelled, 𝑠, the initial velocity, 𝑢, the final velocity, 𝑣, the acceleration, 𝑎, and the time taken, 𝑡.

In this case, we don’t have the time taken, 𝑡. But that doesn’t matter. We need to use the SUVAT equation that deals with these four quantities 𝑢, 𝑣, 𝑎, and 𝑠. And that SUVAT equation happens to be: 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Now, in this case, we’re trying to solve for what the initial velocity is. We’re trying to solve for 𝑢. So we need to rearrange this equation.

First thing we can do is to subtract two 𝑎𝑠 from both sides. This means that two 𝑎𝑠 cancels on the right-hand side, leaving us with 𝑣 squared minus two 𝑎𝑠 is equal to 𝑢 squared. Now, what we can do is to take the square root of both sides of the equation. On the right-hand side, once again, the squared cancels with the square root. And so we find that the square root of 𝑣 squared minus two 𝑎𝑠 is equal to 𝑢.

At this point, all we need to do is to sub in the values of 𝑣, 𝑎, and 𝑠, which looks something like this: 𝑢 is equal to the square root of 𝑣 squared, which is 14 squared, minus two times 𝑎, which is five, times 𝑠, which is 17.1. And since all of the quantities that we’ve used are in their standard units, 𝑣 is in meters per second, 𝑎 is in meters per second squared, and 𝑠 is in meters, the value that we find for 𝑢 is going to be in its standard units. Now, 𝑢 is the initial velocity of the object and the standard unit of velocity is meters per second. So our answer is going to be in meters per second. Plugging this into our calculator, we find that the value of 𝑢 is five meters per second.

And this is our final answer. The object’s initial velocity is five meters per second.