Question Video: Calculating the Acceleration and Time of Flight of an Object Physics

Video Transcript

A bullet fired from a gun leaves the end of the 55-centimeter-long gun barrel at a speed of 500 meters per second and hits a bottle 35 meters away, as shown in the diagram. What is the average acceleration of the bullet in the gun barrel? Round your answer to the nearest kilometer per square second.

Okay, so in this question, we are asked to work out the average acceleration of a bullet fired from a gun. The bullet is only accelerating while it’s in the gun barrel, so the distance over which it accelerates is 55 centimeters. We are told that the bullet leaves the gun at the end of this 55-centimeter distance with a speed of 500 meters per second. Before the gun is fired, we know that the bullet will be at rest, so we can say that it starts with an initial speed of zero meters per second.

We’ll label this initial speed as 𝑢, and we’ll label the bullet’s final speed of 500 meters per second as 𝑣. While we’re at it, let’s label the 55-centimeter length of the gun barrel as 𝑠. So the bullet accelerates from an initial speed of 𝑢 equal to zero meters per second to a final speed of 𝑣 equal to 500 meters per second over a distance of 𝑠 equal to 55 centimeters. We are trying to work out the average acceleration of the bullet. So let’s label this average acceleration as 𝑎.

Now, it turns out that there’s an equation that links an object’s initial speed 𝑢, its final speed 𝑣, its acceleration 𝑎, and the distance 𝑠 over which it accelerates. Specifically, that equation tells us that 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑠. In this case, we know the values of the quantities 𝑣, 𝑢, and 𝑠. We’re trying to work out the acceleration 𝑎. This means that we want to take this equation and rearrange it to make 𝑎 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, which cancels out with the minus 𝑢 squared. This leaves us with an equation that says 𝑣 squared minus 𝑢 squared is equal to two times 𝑎 times 𝑠. Then, we divide both sides of the equation by two 𝑠. On the right-hand side, the two in the numerator cancels with the two in the denominator and the 𝑠 in the numerator cancels with the 𝑠 in the denominator. This gives us an equation that says 𝑣 squared minus 𝑢 squared divided by two 𝑠 is equal to 𝑎. And of course, we can also write this equation the other way around to say that 𝑎 is equal to 𝑣 squared minus 𝑢 squared divided by two 𝑠.

So we now have an equation that will allow us to calculate the value of 𝑎 as long as we know the values of 𝑣, 𝑢, and 𝑠.

If we look back at the question, we see that we are asked to give our answer for the acceleration to the nearest kilometer per square second. However, at the moment our speeds are in units of meters per second and our distance is in units of centimeters. We need to convert the two speeds to units of kilometers per second and the distance to units of kilometers.

In order to do this, we need to recall that one kilometer is equal to 1000 meters and one meter is equal to 100 centimeters. Now, if one kilometer is equal to 1000 meters, then one meter must be equal to one thousandth of a kilometer. Likewise, if one meter is equal to 100 centimeters, then one centimeter is one hundredth of a meter. So to convert the distance 𝑠 of 55 centimeters into units of kilometers, we take the value in units of centimeters and we multiply it by one over 100 meters per centimeter, then again by one over 1000 kilometers per meter.

Looking at the units, we can see that the centimeters cancel with the per centimeter and the meters cancel with the per meter. So we are left with units of kilometers. Evaluating the expression, we find that 𝑠 is equal to 0.00055 kilometers. Now, we just need to convert our speeds to units of kilometers per second. Let’s clear ourselves some space to do this. We’ll start with our speed of 𝑣 equal to 500 meters per second. To get this speed into units of kilometers per second, we need to multiply the value in meters per second by one over 1000 kilometers per meter.

Then, in terms of the units, the meters cancel with the per meter, and we are left with units of kilometers per second. When we evaluate this expression, we find that 𝑣, the speed of the bullet when it leaves the gun barrel, is equal to 0.5 kilometers per second. Now, we could also do the same thing for the initial speed 𝑢. But it’s easier to just notice that since this initial speed is zero meters per second, then it’s going to be zero in whatever units we choose to express it in. After all, if something isn’t moving, then, well, it’s not moving, no matter what units we choose to measure that lack of movement in.

So we know that 𝑢 is equal to zero kilometers per second. And we’ve also worked out that 𝑣 is equal to 0.5 kilometers per second. So we now have values for 𝑠, 𝑢, and 𝑣 in units of kilometers and kilometers per second. All that’s left for us to do is to sub those values into this equation to calculate the value of 𝑎.

Let’s clear some space so that we can do this. We know that 𝑎 is equal to 𝑣 squared minus 𝑢 squared divided by two 𝑠. In place of 𝑣, we’ll sub in our value of 0.5 kilometers per second. And in place of 𝑢, we’ll sub in the value of zero kilometers per second. Finally, in place of the 𝑠 in the denominator, we’ll sub in our value of 0.00055 kilometers. This gives us an expression for 𝑎 and all that we need to do now is to evaluate it.

In the numerator, 0.5 kilometers per second all squared gives us 0.25 kilometers squared per second squared. And zero kilometers per second all squared is zero kilometers squared per second squared. In the denominator, two multiplied by 0.00055 kilometers gives us 0.0011 kilometers. In the numerator of this expression, we have 0.25 kilometers squared per second squared minus zero kilometers squared per second squared. And this is simply equal to 0.25 kilometers squared per second squared.

In terms of the units, one factor of the kilometers from the numerator cancels with the kilometers from the denominator. Then, evaluating the expression gives us that the acceleration 𝑎 is equal to 227.27 kilometers per second squared, where the bar over these two digits indicates that they are recurring. Finally, we should notice that the question asks us to round our answer to the nearest kilometer per square second. So we need to take this value that we’ve calculated for 𝑎 and round it to the nearest whole number of kilometers per second squared. When we do this, we get our answer to this first part of the question that, to the nearest kilometer per square second, the average acceleration of the bullet in the gun barrel is equal to 227 kilometers per second squared.

Okay, now let’s look at the second part of the question.

How much time after the bullet leaves the gun does it hit the bottle, assuming it moves with a constant velocity?

Okay, so we know that the bullet leaves the gun with a speed of 500 meters per second. And we can see from the diagram that the bullet is headed directly toward the bottle. We are told to assume that the bullet moves with a constant velocity. This means that we can assume that both its speed and its direction remain constant. We are told that the bottle is 35 meters away from the gun. So we know that the bullet moves a distance of 35 meters at a speed of 500 meters per second.

We are asked to work out how much time it takes for the bullet to move this 35-meter distance from the gun to the bottle. We’ll label this distance as 𝑑 and the time that we’re trying to calculate as 𝑡, and we’ve already labeled the bullet’s speed as 𝑣. We can recall that these three quantities are linked by the equation 𝑣 is equal to 𝑑 divided by 𝑡. Or in words, for an object moving with a constant speed, that speed is equal to the distance moved by the object divided by the time taken to move the distance.

In this case, we know the values of 𝑣 and 𝑑, and we’re trying to work out the value of the quantity 𝑡. So let’s take this equation and rearrange it to make 𝑡 the subject. To do this, the first step is to multiply both sides of the equation by 𝑡. Then on the right-hand side, the 𝑡 in the numerator cancels with the 𝑡 in the denominator. This leaves us with an equation that says 𝑣 multiplied by 𝑡 is equal to 𝑑. Then, we divide both sides of the equation by 𝑣. On the left-hand side, the 𝑣 in the numerator cancels with the 𝑣 in the denominator.

So we end up with an equation that says time 𝑡 is equal to distance 𝑑 divided by speed 𝑣. Now, we just need to sub our values for 𝑑 and 𝑣 into this equation. When we do this, we get that 𝑡 is equal to 35 meters, that’s our value for 𝑑, divided by 500 meters per second, that’s our value for 𝑣. Evaluating this expression, we find that 𝑡 is equal to 0.07 seconds. And so our answer to this second part of the question is that the bullet hits the bottle 0.07 seconds after it leaves the gun.