### Video Transcript

An object with a velocity ๐ฃ is slowed to rest by a constant force ๐น over a distance of 12 meters. If the objectโs velocity is increased to five ๐ฃ and then ๐น is again applied to decelerate the object, how far does the object move between the force being applied and the object coming to rest?

Okay, so in this question, we know that weโve got an object which initially has a velocity ๐ฃ, which is then slowed to rest by a constant force ๐น. And it slows to rest over a distance of 12 meters. If we then take that same object and change its initial velocity to five ๐ฃ and then use the same force ๐น to decelerate the object, we need to find out how far the object moves between the force being applied and the object coming to rest.

So basically, weโve got two scenarios where the object that weโve got is decelerating. The first scenario is where its initial velocity is ๐ฃ and the second scenario is when its initial velocity is five ๐ฃ. Letโs draw a diagram of both scenarios.

First of all, letโs say that this is our object โ just a regular ball with a mass ๐. We donโt actually know what this mass is, but as weโll see later we donโt need to know. Now, in the first scenario, this object is starting out with a velocity ๐ฃ. So letโs say itโs moving towards the right. And we know it travels some distance and is decelerating the whole time until it stops. The distance it travels while decelerating until it stops has been given as 12 meters. Letโs call this distance ๐ . Now whatโs slowing the object down is a force that we apply and this force must be in the opposite direction because itโs decelerating the object. Itโs slowing it down. So thatโs the first situation.

Now, letโs draw the second scenario. This time weโve got the same object again. This ball with a mass of ๐. But it starts out with a velocity of five ๐ฃ. And once again, we apply the same force to it ๐น so that it decelerates over some distance until it stops. Now, weโre asked to work out what this distance is. So letโs call this distance ๐ฅ.

Now, in both cases, the object is initially moving. Therefore, it must have some kinetic energy. We can recall that the kinetic energy of an object ๐ธ sub ๐ is given by multiplying half by ๐ โ the mass of the object โ by ๐ฃ โ the velocity of the object โ squared. So using this equation, we can work out in both cases the change in kinetic energy of the object as it slows to a stop. Letโs call the change in kinetic energy ฮ ๐ธ sub ๐ because we use ฮ to represent a change and ๐ธ sub ๐ represents our kinetic energy.

Now, in these two scenarios, the change in kinetic energy is not going to be the same. So letโs say for the first one our change in kinetic energy is ฮ ๐ธ sub ๐ comma one. And for the second scenario, weโve got ฮ ๐ธ sub ๐ comma two. Now, the change in kinetic energy ฮ ๐ธ sub ๐ in general for both cases is going to be the final kinetic energy ๐ธ sub ๐ final minus the initial kinetic energy ฮ ๐ธ sub ๐ initial.

So for the first scenario ฮ ๐ธ sub ๐ comma one, the final kinetic energy is half multiplied by the mass of the object multiplied by the final velocity of the object which is zero because the object is slowing to rest. And so this time becomes zero because weโve got something multiplied by zero and thatโs zero. So thatโs the final kinetic energy.

And now, we can work out the initial kinetic energy. Thatโs simply becomes half multiplied by the mass of the object multiplied by the initial velocity squared. Now, the initial velocity is ๐ฃ. So we got half ๐๐ฃ squared. And so the whole expression becomes zero minus half ๐๐ฃ squared, which is simply negative one half ๐๐ฃ squared. So thatโs ฮ ๐ธ sub ๐ comma one.

We can do the same for ฮ ๐ธ sub ๐ comma two. Now, the final kinetic energy once again is going to be zero because even in this situation the object is coming to rest. And the final kinetic energy is going to be, and the initial kinetic energy is going to be half multiplied by ๐ multiplied by the initial velocity, which in this case is five ๐ฃ squared. Now, when we square five ๐ฃ, we get 25 ๐ฃ squared. Now here what we can do is to pull out the factor of 25. So weโre left with 25 times half ๐๐ฃ squared. And the reason weโve done that is because half ๐๐ฃ squared kind of looks like this. And so overall, ฮ ๐ธ sub ๐ comma two is equal to negative 25 times half ๐๐ฃ squared.

If we wanted to, we could have also taken this negative sign in here so that the bit inside the parenthesis looks identical to ฮ ๐ธ sub ๐ comma one. In fact, yeah, letโs do that. Now weโve got ฮ ๐ธ sub ๐ comma two is equal to 25 times negative half ๐๐ฃ squared. This will come in handy later. For now though, letโs look at some other information.

We know that in both cases weโre using a force ๐น to decelerate the object. We can use this information to work out the amount of work done on the object in order to slow it down. We can recall that the amount of work done on an object ๐ค is equal to the force applied to the object ๐น multiplied by the distance travelled by the object ๐. So we can calculate the amount of work done in the first scenario as ๐ค is equal to ๐น the force multiplied by the distance ๐ . And letโs call this ๐ค sub one for the work done in the first scenario. ๐ค sub two then becomes ๐น multiplied by the distance travelled which is ๐ฅ and this ๐ฅ is what weโre trying to find out here.

So why is this relevant? Well, we can use something known as the work energy theorem. The work energy theorem tells us that the amount of work done on an object is equal to the change in kinetic energy of that object. In other words, ๐ค is equal to ฮ ๐ธ sub ๐. So in each scenario, we can equate ๐ค with ฮ ๐ธ sub ๐.

So letโs do that for scenario one first. Negative one-half ๐๐ฃ squared โ thatโs ฮ ๐ธ sub ๐ comma one โ is equal to ๐น๐ . Thatโs ๐ค sub one. Now hang on, weโve got a negative on one side, but a positive on the other? Well, no, we donโt. Remember the force is acting against the direction of motion. So the value of force is negative. And hence, weโve got two negative values on each side of the equation.

Anyway, now letโs do the same thing for the second scenario. What we have here is 25 times negative one-half ๐๐ฃ squared is equal to ๐น๐ฅ. And at this point, we can see why itโs useful to have written the left-hand side as 25 times negative half ๐๐ฃ squared. Because negative one-half ๐๐ฃ squared is equal to ๐น๐ , so we can substitute that into here. So we have 25 times ๐น๐ is equal to ๐น๐ฅ. Letโs now divide both sides of the equation by ๐น so that all of these ๐นs cancel out. And what weโre left with is 25๐ is equal to ๐ฅ.

And remember in this equation, we weโre trying to find out what the value of ๐ฅ is. And now weโve got it in terms of ๐ , just the distance travelled in the first scenario. So we can calculate what ๐ฅ is. ๐ฅ is equal to 25 times ๐ which happens to be 12 meters which ends up being 300 meters.

And so we find out that our final answer is that the object moves 300 meters between the force first being applied and the object coming to rest.