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Question Video: Acceleration over a Distance Physics • 9th Grade

An object starts from rest and accelerates at a rate of 2 m/s² along a straight line until its velocity reaches 8 m/s. How far has the object traveled from its starting position when it reaches this velocity?

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Video Transcript

An object starts from rest and accelerates at a rate of two meters per second squared along a straight line until its velocity reaches eight meters per second. How far has the object traveled from its starting position when it reaches this velocity?

Okay, so in this question, we have an object that we’re told starts from rest. This means that the object’s initial velocity, which we’ve labeled as 𝑢, has a value of zero meters per second. We’re told that the object accelerates until its velocity reaches eight meters per second. So we know that the object’s final velocity, which we’ve labeled as 𝑣, has a value of eight meters per second. The question tells us that the object accelerates at a rate of two meters per second squared. Let’s label this acceleration as 𝑎. So, what we know is that the object starts at some position with an initial velocity of zero meters per second and that after accelerating along a straight line at a rate of two meters per second squared, the object ends up with a final velocity of eight meters per second. And this final velocity must have the same direction as the acceleration.

We’re asked to work out how far the object has traveled from its starting position when it reaches this final velocity. So that’s the distance traveled from this position to this position here. Let’s label this distance traveled as 𝑠. And then, to answer this question, what we need to do is to find the value of 𝑠. Now, we can recall that there is an equation that links the four quantities 𝑢, 𝑣, 𝑎, and 𝑠. Specifically, that equation is this one here, which tells us that 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑠. In order for this equation to apply, the acceleration must be a constant value. In this case, we’re told the acceleration is two meters per second squared, which is indeed a constant.

As well as this, for the equation to apply, the object’s motion must be in a straight line. Luckily, in this case, we’re told in the question that the object is moving along a straight line. So we can safely use this equation. Now, we’re trying to find the value of the quantity 𝑠. So we need to rearrange this equation to make 𝑠 the subject. The first step is to subtract 𝑢 squared from both sides of the equation. Then, on the right-hand side, we have 𝑢 squared and minus 𝑢 squared. So these two terms cancel each other out. This leaves us with 𝑣 squared minus 𝑢 squared is equal to two times 𝑎 times 𝑠. Then, we can divide both sides of the equation by two 𝑎.

On the right-hand side, the two 𝑎 in the numerator cancels with the two 𝑎 in the denominator. And so we have that 𝑣 squared minus 𝑢 squared divided by two 𝑎 is equal to 𝑠. And of course, we can also write this the other way around to say that 𝑠 is equal to 𝑣 squared minus 𝑢 squared divided by two 𝑎. All that’s left for us to do is to substitute these values into this equation. Substituting in that our final velocity 𝑣 is equal to eight meters per second, the initial velocity 𝑢 is equal to zero meters per second, and the acceleration 𝑎 is equal to two meters per second squared, we have that 𝑠 is equal to the square of eight meters per second minus the square of zero meters per second all divided by two times two meters per second squared.

Evaluating each of the terms on the right-hand side, we have that 𝑠 is equal to 64 meters squared per second squared minus zero meters squared per second squared divided by four meters per second squared. Now, 64 meters squared per second squared minus zero meters squared per second squared is simply 64 meters squared per second squared. So then, the last step is to evaluate the result of this division. In terms of the units, the per second squared in the numerator cancels with the per second squared in the denominator. And one factor of meters from the numerator cancels with the meters from the denominator. This leaves us with units of meters, which makes sense since 𝑠 is a distance.

When we divide 64 by four, we get a result of 16. And so our answer to the question of “How far the object has traveled from its starting position when it reaches the final velocity of eight meters per second?” is that the object has traveled 16 meters.

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