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

Question Video: Acceleration over a Distance Physics • First Year of Secondary School

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An object has an initial velocity of 3 m/s and accelerates in the direction of its velocity along a straight line at a rate of 4 m/s². Its velocity reaches 11 m/s when it is at the end of the line. What is the length of the line?

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Video Transcript

An object has an initial velocity of three metres per second and accelerates in the direction of its velocity along a straight line at a rate of four metres per second squared. Its velocity reaches 11 metres per second when it is at the end of the line. What is the length of the line?

Okay, so in this question, we’ve got an object. Let’s say it starts here. And let’s say it’s moving in this direction to the right with its initial velocity which we’ve been told is three metres per second. Now, we’ve been told that it accelerates in the direction of its velocity along a straight line at a rate of four metres per second squared. So the object is travelling in a straight line towards the right, as we’ve drawn it. And by the time it gets to the end of the line, we’ve been told that it’s moving at 11 metres per second. And we’ve been asked to find the length of this line, from the start position of the object to the finish position.

Let’s say that the length of this line is 𝑠. And 𝑠 is what we’re trying to find. As well as this, let’s label the initial velocity of the object as 𝑢, three metres per second. Let’s say the final velocity of the object is 𝑣, 11 metres per second. And let’s say that the acceleration of the object is 𝑎, four metres per second squared. Now, the equation that links 𝑢, 𝑣, 𝑎, and 𝑠 is this one here: 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Or, the final velocity of the object squared is equal to its initial velocity squared plus two times its acceleration times the distance it travels.

Now, it’s important to know that for this equation to apply, the acceleration must be a constant value. And in this case, we’ve been told that it’s four metres per second squared which is a constant. As well as this, the distance that the object travels in order for this equation to apply must be in a straight line. And this is the case as we’ve been told in the question. Therefore, we can use this equation to solve our problem. We already know the values of 𝑣, 𝑢, and 𝑎. All we need to do is to rearrange to solve for 𝑠. We can start by subtracting 𝑢 squared from both sides of the equation. This way, it cancels on the right-hand side. And what we’re left with is 𝑣 squared minus 𝑢 squared on the left and two 𝑎𝑠 on the right. Then, we can divide both sides of the equation by two 𝑎. This way the two on the right cancels and the 𝑎 on the right cancels. And we have 𝑣 squared minus 𝑢 squared divided by two 𝑎 is equal to 𝑠.

So at this point, all we need to do is to plug in the values. 𝑠 is equal to 𝑣, which is the final velocity, squared, so 11 metres per second squared, minus 𝑢, which is the initial velocity, squared, which is three metres per second whole squared, divided by two times the acceleration, which is four metres per second squared.

Expanding the parentheses in the numerator, we have 121 metres squared per second squared minus nine metres squared per second squared. And so, the numerator becomes 112 metres squared per second squared. Then, we divide this by two times four, which is eight, metres per second squared. And so, the per second squared in the numerator and the denominator cancel. And one factor of metres in the numerator cancels with the one factor of metres in the denominator. Overall, we’re just left with a unit of metres, which is good because we’re calculating a distance which has units of metres. And hence, we’re left with the final answer to our question. The length of the line along which the object moves is 14 metres.

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