### Video Transcript

An object has a mass of nine grams. The mass of this object is reduced with a rate of one gram per second. The velocity of the object is given by π― equals three π‘ plus 10π’ centimeters per second. Find the magnitude of the force acting on this object at π‘ equals two.

Now, initially, weβre given information about the mass of the object and the fact that the mass of this object reduces over time. Now, we normally look to find the magnitude of a force given its mass and acceleration, where force is mass times acceleration. However, when given an object with variable mass, we use an alternative formula. That is, for an object with mass π and velocity π£, the force is given by π times dπ£ by dπ‘ plus π£ times dπ by dπ‘.

Now, the question gives us an expression for π£ at time π‘. So we can quite easily calculate dπ£ by dπ‘. But how do we calculate π and dπ by dπ‘? Well, since the mass of the object reduces at a rate of one gram per second, we can say that the change in mass with respect to time, that is, dπ by dπ‘, is equal to negative one. Then, we can integrate this expression with respect to π‘ and solve to find the constant of integration to find an expression for π. In other words, π is equal to the indefinite integral of negative one with respect to π‘.

Now, the integral of a constant is just the value of that constant multiplied by the variable. So here the integral of negative one is negative π‘. So π is equal to negative π‘ plus some constant of integration πΆ. But in fact weβre told information about the initial mass of the object. At π‘ equals zero, the mass of the object is nine grams. So we can rewrite our equation substituting these values in. That is, nine equals negative zero plus πΆ, meaning πΆ is equal to nine. So our expression for mass of the object at time π‘ is π equals negative π‘ plus nine.

And we might notice that we do need to restrict the domain of this function. π‘ will be able to take values less than or equal to nine. And this is because if we substitute anything greater than nine into the expression, weβll end up with a negative value for the mass. Now, in fact, this doesnβt really matter. Weβre finding the magnitude of the force at π‘ equals two, but itβs worth bearing in mind. So we have our expression for π and dπ by dπ‘. And then we have a vector equation for π£. Itβs π― equals three π‘ plus 10π’.

Now, itβs actually really useful to spot that since three π‘ plus 10 is the π’-component of this vector, the velocity of the object is only acting in a straight line. And so we can actually identify that the magnitude of the velocity will be three π‘ plus 10 for our value of π‘. Next, we need to calculate an expression for dπ£ by dπ‘. Well, the derivative of three π‘ with respect to π‘ is three, and the derivative of 10 is zero. So dπ£ by dπ‘ is equal to three. This essentially tells us that the acceleration of the object is constant. It accelerates at a rate of three centimeters per second per second.

And now we can substitute everything we know about our object into the given formula. Itβs πΉ equals π, which is negative π‘ plus nine, times dπ£ by dπ‘, which is three, plus π£, which is three π‘ plus 10, times dπ by dπ‘, thatβs negative one. Letβs distribute our parentheses. That gives us πΉ equals negative three π‘ plus 27 minus three π‘ minus 10, which is negative six π‘ plus 17. So we have an expression for the magnitude of the force at a given time π‘.

Now, the question asks us to find the magnitude of the force at π‘ equals two. So thatβs πΉ equals negative six times two plus 17. Thatβs negative 12 plus 17, which is equal to five. Now, since weβre working in grams, centimeters, and second, our unit for the magnitude of the force is dynes. And so the magnitude of the force acting on the object at π‘ equals two is five dynes.

Now, itβs worth noting that we calculated the velocity in terms of its magnitude. And we were able to do this because the object is traveling in a straight line. We could alternatively have used the vector form of our velocity. We could have differentiated this with respect to π‘. And we would have found that the derivative of the vector π‘ with respect to π‘ is three π. Ultimately, that wouldβve given us a vector expression for πΉ from which we then couldβve calculated the magnitude of the force. Either way, we get πΉ equals five dynes.