# Video: Determine the Force Acting upon a Body That Moves in a Straight Line and Whose Mass Varies with Time

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠 = (2𝑡² + 5𝑡 + 4) m. Its mass varies with time such that 𝑚 = (6𝑡 + 5) kg. Determine the force acting upon the body when 𝑡 = 3𝑠.

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### Video Transcript

A body moves in a straight line. At time 𝑡 seconds, its displacement from a fixed point is given by 𝑠 equals two 𝑡 squared plus five 𝑡 plus four meters. Its mass varies with time such that 𝑚 is equal to six 𝑡 plus five kilograms. Determine the force acting upon the body when 𝑡 equals three seconds.

In this question, we’ve been given information about the displacement of a body moving in a straight line with respect to time. We’ve also been given information about its mass with respect to time 𝑡. And so we have a body with variable mass. This means we can’t use the equation 𝐹 equals 𝑚𝑎 that we’re so used to.

Instead, we use Newton’s second law, but we add a term to account for the momentum carried by mass entering the system. We say 𝐹 is equal to 𝑣 times d𝑚 by d𝑡 — that’s the derivative of mass with respect to time — plus 𝑚 by d𝑣 by d𝑡, which is the derivative of 𝑣 with respect to time, in other words its acceleration.

Now since 𝑚 is equal to six 𝑡 plus five, we’re going to need to differentiate that with respect to time. And we can do that term by term. The derivative of six 𝑡 is just six, and the derivative of five is zero. So we so find that d𝑚 by d𝑡 is equal to six.

Now we’re told that 𝑠, the displacement, is two 𝑡 squared plus five 𝑡 plus four. And so we need to recall that velocity is the change in displacement with respect to time. That’s d𝑠 by d𝑡. So we differentiate each term with respect to time. The derivative of two 𝑡 squared is two times two 𝑡, which is four 𝑡. And the derivative of five 𝑡 is five. The derivative of four, however, is zero. So we find when we differentiate 𝑠 with respect to 𝑡, we get four 𝑡 plus five. Similarly, we need d𝑣 by d𝑡 for the formula. So we’re going to differentiate once again. And of course, in fact, that gives us acceleration.

Now the derivative of four 𝑡 is four, and the derivative of five is zero. Let’s substitute everything we have into our formula. 𝑣 times d𝑚 by d𝑡 is four 𝑡 plus five times six, and 𝑚 times d𝑣 by d𝑡, in other words 𝑚𝑎, is six 𝑡 plus five times four. We distribute each of our parentheses. Four 𝑡 times six is 24𝑡, and five times six is 30. Similarly, six 𝑡 times four is 24𝑡, and five times four is 20. And so we collect like terms, and we find that the force is 48𝑡 plus 50. And since we’re working meters, seconds, and kilograms, it’s actually 48𝑡 plus 50 newtons.

Now, of course, we’re looking to find the exact force at 𝑡 equals three. So 𝐹 equals 48 times three. We replace 𝑡 with three and then we add 50. That gives us 194. And so the force acting on the body when 𝑡 equals three seconds is 194 newtons.