### Video Transcript

The momentum of an object is given by the formula momentum equals mass times velocity, expressed in symbols as π equals ππ£. The velocity of an object is given by the formula velocity equals displacement over time, expressed in symbols as π£ equals π over π‘. Which of the following is a correct formula? A) ππ£ equals ππ , B) ππ£ equals ππ‘, C) ππ equals ππ‘, D) ππ‘ equals π£π , or E) ππ‘ equals ππ .

To answer this question, weβll need to look at each answer option in turn and decide whether itβs correct or incorrect. A good method for doing this is to start with the left-hand side from each answer option and try to match the right-hand side of the expression by substituting and rearranging the equations given in the question. This means weβll be looking for expressions with ππ£ on the left-hand side, with ππ on the left-hand side, and with ππ‘ on the left-hand side.

Itβs important to notice that each of the answer options contains symbols from the equations π equals ππ£ and π£ equals π over π‘. This means weβll need to combine these two equations together in order to find the expressions that weβre looking for. So, first, letβs find an expression where the left-hand side is π times π£.

Letβs start just by writing down the left-hand side of the equation that we want. Then, letβs think about how we can use the two equations weβve been given to come up with a right-hand side for our formula. Equation number one tells us that π is equivalent to π times π£. This means that we can replace the π in our expression with ππ£. So, π times π£ becomes ππ£ times π£, which we can simplify to ππ£ squared. This expression is correct. But weβre not done yet.

Because we want to find out whether one of these expressions is correct, then weβre looking for a formula that not only has ππ£ on the left-hand side, but also contains the variables π or π‘ on the right-hand side. The way we can achieve this is by substituting equation two into the right-hand side of our expression. That way, weβll express any π£s in terms of π and π‘. So, we can just replace this π£ with π over π‘. Notice that weβve used parentheses here just to keep things clear.

Because π£ is equal to π over π‘, that means π£ squared is equal to π over π‘ squared. So, we need to make sure that we square the entire fraction that weβve substituted in. When squaring any fraction, we square the numerator, and we square the denominator. So, here, that gives us π times π squared over π‘ squared. Okay, so weβve now found an expression with ππ£ on the left and the desired variables on the right. But if we look at A and B, weβll see that those expressions only contain π or π‘ on the right-hand side and not both.

Now, we could attempt to keep substituting this equation into our expression in order to express the right-hand side in terms of just π or just π‘. But weβd quickly find that we canβt do this without also introducing extra factors of π£, which arenβt present in our answer options. So, we know that neither A nor B are the correct answer.

Next, letβs find a formula with ππ on the left-hand side. And again, weβll start just by writing down the left-hand side of the equation that we want. And just like before, we can start by substituting π equals ππ£ into our expression. So, ππ becomes ππ£π . Notice that this gives us an expression with π on the right-hand side, just like in our answer option. However, if we want to see if this option is correct, then we want to try to express the right-hand side in terms of π and π‘.

Once again, substituting equation two into our expression can help us. Substituting π over π‘ in place of π£ gives us π times π over π‘ times π , which we can simplify to give us ππ squared over π‘. And we can stop there because weβve managed to express ππ in terms of π and π‘. But itβs not the same as option C. Then, we know that option C is incorrect as well.

That leaves us with just two possible answers, ππ‘ equals π£π and ππ‘ equals ππ . To determine which of these is correct, weβll need to come up with an expression that has ππ‘ on the left-hand side. So, once again, weβll start by just writing down the left-hand side of the equation that we want. And again, we can use equation one to substitute ππ£ in place of π, which changes ππ‘ into ππ£π‘.

Here, the possible answers both have π on the right-hand side. So, we want to make a substitution that will give us π on the right-hand side of our equation. And once again, we can do that by using equation two, which allows us to replace π£ with π over π‘. That gives us π times π over π‘ times π‘, which we can simplify to give us ππ π‘ divided by π‘.

Our final step here is to simplify this fraction. Because there is a factor of π‘ in the numerator and in the denominator, they effectively cancel each other out. Leaving us with the expression ππ‘ equals ππ , which is the same as Option E. So, we know that D is incorrect as well. And the correct answer is option E, ππ‘ equals ππ .